Sum of all values that satisfy $fracx^x-3x=fracxx^frac4x$. The 2019 Stack Overflow Developer Survey Results Are InSum of all real numbers $x$ such that $(textA quadratic)^textAnother quadratic=1$.Prove the following equation, given that $a,b,c,d$ are reals such thatWhen can you treat a limit like an equation?Solving $3x - 40 cdot 2 = x + 5$Finding the inverse function of $f(x) = 5 - 4x^3$I get two different answers on simple equation. What am I doing wrong?What is the sum of all integer values of $x$ such that $(x^2-17x+71)^(x^2-34x+240)=1$?Solve for $x$ the equation $ x^log_10 x=fracx^3100$If $ a,b,c$ are in descending order of magnitude how to prove $(fraca+ca-c)^a <(fracb+cb-c)^b$if $m<frac12$, what is $-frac1m$ equal to?Why is the solution to the equation $ln(x)+ln(3x+1) = 0$ not $x=frac-14$?
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Sum of all values that satisfy $fracx^x-3x=fracxx^frac4x$.
The 2019 Stack Overflow Developer Survey Results Are InSum of all real numbers $x$ such that $(textA quadratic)^textAnother quadratic=1$.Prove the following equation, given that $a,b,c,d$ are reals such thatWhen can you treat a limit like an equation?Solving $3x - 40 cdot 2 = x + 5$Finding the inverse function of $f(x) = 5 - 4x^3$I get two different answers on simple equation. What am I doing wrong?What is the sum of all integer values of $x$ such that $(x^2-17x+71)^(x^2-34x+240)=1$?Solve for $x$ the equation $ x^log_10 x=fracx^3100$If $ a,b,c$ are in descending order of magnitude how to prove $(fraca+ca-c)^a <(fracb+cb-c)^b$if $m<frac12$, what is $-frac1m$ equal to?Why is the solution to the equation $ln(x)+ln(3x+1) = 0$ not $x=frac-14$?
$begingroup$
What is the sum of all values of $x$ that satisfy the equation
$fracx^x-3x=fracxx^frac4x$?
I start off by cross multiplying.
$$x^2=x^x-3+frac4x$$
Taking the square root of both sides gives me:
$$x=pm x^frac12x-frac32+frac42x$$
I start with the positive side first:
$$1=frac12x-frac32+frac42x$$$$x^2-5x+4=0$$$$x=4, 1$$
Now, I start with the negative side:
$$x=-x^frac12x-frac32+frac42x$$$$x^fracx^2-5x+42x=-1$$
I can't log both sides because of the $-1$, so I raise both sides by a power of $2x$ to get rid of the fraction. Because $2x$ is even, the RHS becomes a $1$.
$$x^x^2-5x+4=1$$
Now, taking the log of both sides, I have
$$(x^2-5x+4)cdotoperatornamelog(x)=0$$
Dividing both sides by $operatornamelog(x)$, I get $x^2-5x+4=0$ again, which should give me $x=4, 1$ again. So thus, the answer should be $4+1=5$.
However, this is wrong, as the answer key says that the answer is $-1+1+4=boxed4$. I have checked $-1$ as a solution on all of my equations, and all of them work. How can I derive the $-1$ out from my equations? Or, if my approach or any of my equations are wrong, how do I get the answer $x=-1$?
algebra-precalculus
edited Mar 24 at 7:21
NoChance
3,76621321
asked Mar 24 at 5:39
Max0815Max0815
81418
$endgroup$
add a comment |
$begingroup$
What is the sum of all values of $x$ that satisfy the equation
$fracx^x-3x=fracxx^frac4x$?
I start off by cross multiplying.
$$x^2=x^x-3+frac4x$$
Taking the square root of both sides gives me:
$$x=pm x^frac12x-frac32+frac42x$$
I start with the positive side first:
$$1=frac12x-frac32+frac42x$$$$x^2-5x+4=0$$$$x=4, 1$$
Now, I start with the negative side:
$$x=-x^frac12x-frac32+frac42x$$$$x^fracx^2-5x+42x=-1$$
I can't log both sides because of the $-1$, so I raise both sides by a power of $2x$ to get rid of the fraction. Because $2x$ is even, the RHS becomes a $1$.
$$x^x^2-5x+4=1$$
Now, taking the log of both sides, I have
$$(x^2-5x+4)cdotoperatornamelog(x)=0$$
Dividing both sides by $operatornamelog(x)$, I get $x^2-5x+4=0$ again, which should give me $x=4, 1$ again. So thus, the answer should be $4+1=5$.
However, this is wrong, as the answer key says that the answer is $-1+1+4=boxed4$. I have checked $-1$ as a solution on all of my equations, and all of them work. How can I derive the $-1$ out from my equations? Or, if my approach or any of my equations are wrong, how do I get the answer $x=-1$?
algebra-precalculus
edited Mar 24 at 7:21
NoChance
3,76621321
asked Mar 24 at 5:39
Max0815Max0815
81418
$endgroup$
1
$begingroup$
See math.stackexchange.com/questions/3157637/…
$endgroup$
– lab bhattacharjee
Mar 24 at 6:13
$begingroup$
This is a good question. In fact if you draw the 2 curves, x=-1 does not show as an intersection point !!!
$endgroup$
– NoChance
Mar 24 at 7:22
$begingroup$
Technically it does. While continuous curves can be rendered only in the first quadrant, the functions over $mathbbR$ also include other individual points where they may be defined. One such point for both functions is $x=-1$, and at that value of $x$ the points are the same for the two functions.
$endgroup$
– Oscar Lanzi
Mar 24 at 11:18
$begingroup$
@OscarLanzi, thank your for your comment. I am still not clear why would a program not plot the said curves at x=-1 if the function is defined at x=-1?
$endgroup$
– NoChance
Mar 24 at 13:48
$begingroup$
@OscarLanzi, thank your for your comment. I was not aware of the relationship between zeroes, algebraic operations and continuity of the function....
$endgroup$
– NoChance
Mar 24 at 14:27
add a comment |
$begingroup$
What is the sum of all values of $x$ that satisfy the equation
$fracx^x-3x=fracxx^frac4x$?
I start off by cross multiplying.
$$x^2=x^x-3+frac4x$$
Taking the square root of both sides gives me:
$$x=pm x^frac12x-frac32+frac42x$$
I start with the positive side first:
$$1=frac12x-frac32+frac42x$$$$x^2-5x+4=0$$$$x=4, 1$$
Now, I start with the negative side:
$$x=-x^frac12x-frac32+frac42x$$$$x^fracx^2-5x+42x=-1$$
I can't log both sides because of the $-1$, so I raise both sides by a power of $2x$ to get rid of the fraction. Because $2x$ is even, the RHS becomes a $1$.
$$x^x^2-5x+4=1$$
Now, taking the log of both sides, I have
$$(x^2-5x+4)cdotoperatornamelog(x)=0$$
Dividing both sides by $operatornamelog(x)$, I get $x^2-5x+4=0$ again, which should give me $x=4, 1$ again. So thus, the answer should be $4+1=5$.
However, this is wrong, as the answer key says that the answer is $-1+1+4=boxed4$. I have checked $-1$ as a solution on all of my equations, and all of them work. How can I derive the $-1$ out from my equations? Or, if my approach or any of my equations are wrong, how do I get the answer $x=-1$?
algebra-precalculus
edited Mar 24 at 7:21
NoChance
3,76621321
asked Mar 24 at 5:39
Max0815Max0815
81418
$endgroup$
What is the sum of all values of $x$ that satisfy the equation
$fracx^x-3x=fracxx^frac4x$?
I start off by cross multiplying.
$$x^2=x^x-3+frac4x$$
Taking the square root of both sides gives me:
$$x=pm x^frac12x-frac32+frac42x$$
I start with the positive side first:
$$1=frac12x-frac32+frac42x$$$$x^2-5x+4=0$$$$x=4, 1$$
Now, I start with the negative side:
$$x=-x^frac12x-frac32+frac42x$$$$x^fracx^2-5x+42x=-1$$
I can't log both sides because of the $-1$, so I raise both sides by a power of $2x$ to get rid of the fraction. Because $2x$ is even, the RHS becomes a $1$.
$$x^x^2-5x+4=1$$
Now, taking the log of both sides, I have
$$(x^2-5x+4)cdotoperatornamelog(x)=0$$
Dividing both sides by $operatornamelog(x)$, I get $x^2-5x+4=0$ again, which should give me $x=4, 1$ again. So thus, the answer should be $4+1=5$.
However, this is wrong, as the answer key says that the answer is $-1+1+4=boxed4$. I have checked $-1$ as a solution on all of my equations, and all of them work. How can I derive the $-1$ out from my equations? Or, if my approach or any of my equations are wrong, how do I get the answer $x=-1$?
algebra-precalculus
algebra-precalculus
edited Mar 24 at 7:21
NoChance
3,76621321
asked Mar 24 at 5:39
Max0815Max0815
81418
edited Mar 24 at 7:21
NoChance
3,76621321
asked Mar 24 at 5:39
Max0815Max0815
81418
edited Mar 24 at 7:21
NoChance
3,76621321
edited Mar 24 at 7:21
NoChance
3,76621321
edited Mar 24 at 7:21
NoChance
3,76621321
3,76621321
asked Mar 24 at 5:39
Max0815Max0815
81418
asked Mar 24 at 5:39
Max0815Max0815
81418
asked Mar 24 at 5:39
Max0815Max0815
81418
81418
1
$begingroup$
See math.stackexchange.com/questions/3157637/…
$endgroup$
– lab bhattacharjee
Mar 24 at 6:13
$begingroup$
This is a good question. In fact if you draw the 2 curves, x=-1 does not show as an intersection point !!!
$endgroup$
– NoChance
Mar 24 at 7:22
$begingroup$
Technically it does. While continuous curves can be rendered only in the first quadrant, the functions over $mathbbR$ also include other individual points where they may be defined. One such point for both functions is $x=-1$, and at that value of $x$ the points are the same for the two functions.
$endgroup$
– Oscar Lanzi
Mar 24 at 11:18
$begingroup$
@OscarLanzi, thank your for your comment. I am still not clear why would a program not plot the said curves at x=-1 if the function is defined at x=-1?
$endgroup$
– NoChance
Mar 24 at 13:48
$begingroup$
@OscarLanzi, thank your for your comment. I was not aware of the relationship between zeroes, algebraic operations and continuity of the function....
$endgroup$
– NoChance
Mar 24 at 14:27
add a comment |
1
$begingroup$
See math.stackexchange.com/questions/3157637/…
$endgroup$
– lab bhattacharjee
Mar 24 at 6:13
$begingroup$
This is a good question. In fact if you draw the 2 curves, x=-1 does not show as an intersection point !!!
$endgroup$
– NoChance
Mar 24 at 7:22
$begingroup$
Technically it does. While continuous curves can be rendered only in the first quadrant, the functions over $mathbbR$ also include other individual points where they may be defined. One such point for both functions is $x=-1$, and at that value of $x$ the points are the same for the two functions.
$endgroup$
– Oscar Lanzi
Mar 24 at 11:18
$begingroup$
@OscarLanzi, thank your for your comment. I am still not clear why would a program not plot the said curves at x=-1 if the function is defined at x=-1?
$endgroup$
– NoChance
Mar 24 at 13:48
$begingroup$
@OscarLanzi, thank your for your comment. I was not aware of the relationship between zeroes, algebraic operations and continuity of the function....
$endgroup$
– NoChance
Mar 24 at 14:27
1
1
$begingroup$
See math.stackexchange.com/questions/3157637/…
$endgroup$
– lab bhattacharjee
Mar 24 at 6:13
$begingroup$
See math.stackexchange.com/questions/3157637/…
$endgroup$
– lab bhattacharjee
Mar 24 at 6:13
$begingroup$
This is a good question. In fact if you draw the 2 curves, x=-1 does not show as an intersection point !!!
$endgroup$
– NoChance
Mar 24 at 7:22
$begingroup$
This is a good question. In fact if you draw the 2 curves, x=-1 does not show as an intersection point !!!
$endgroup$
– NoChance
Mar 24 at 7:22
$begingroup$
Technically it does. While continuous curves can be rendered only in the first quadrant, the functions over $mathbbR$ also include other individual points where they may be defined. One such point for both functions is $x=-1$, and at that value of $x$ the points are the same for the two functions.
$endgroup$
– Oscar Lanzi
Mar 24 at 11:18
$begingroup$
Technically it does. While continuous curves can be rendered only in the first quadrant, the functions over $mathbbR$ also include other individual points where they may be defined. One such point for both functions is $x=-1$, and at that value of $x$ the points are the same for the two functions.
$endgroup$
– Oscar Lanzi
Mar 24 at 11:18
$begingroup$
@OscarLanzi, thank your for your comment. I am still not clear why would a program not plot the said curves at x=-1 if the function is defined at x=-1?
$endgroup$
– NoChance
Mar 24 at 13:48
$begingroup$
@OscarLanzi, thank your for your comment. I am still not clear why would a program not plot the said curves at x=-1 if the function is defined at x=-1?
$endgroup$
– NoChance
Mar 24 at 13:48
$begingroup$
@OscarLanzi, thank your for your comment. I was not aware of the relationship between zeroes, algebraic operations and continuity of the function....
$endgroup$
– NoChance
Mar 24 at 14:27
$begingroup$
@OscarLanzi, thank your for your comment. I was not aware of the relationship between zeroes, algebraic operations and continuity of the function....
$endgroup$
– NoChance
Mar 24 at 14:27
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
The reasoning here is faulty:
I can't log both sides because of the $-1$, so I raise both sides by a power of $2x$ to get rid of the fraction. Because $2x$ is even, the RHS becomes a $1$.
For one thing, what if $x$ is not an integer, and $2x$ is not even?
Starting from $x^fracx^2-5x+42x=-1$, you could absolute value both sides: $$lvert xrvert^fracx^2-5x+42x=1$$
and then take the logarithm: $$fracx^2-5x+42xlnlvert xrvert=0$$ from which you conclude either $x=4$, $x=1$, or $lvert xrvert=1$. Check that $4$, $1$, and $-1$ satisfy the original equation, and you have a complete solution.
answered Mar 24 at 5:53
alex.jordanalex.jordan
39.7k660122
$endgroup$
$begingroup$
Hey, a quick question. When you absolute value both sides, why is only the $x$ absolute valued in the LHS, shouldn't it be the whole LHS? Could you explain why?
$endgroup$
– Max0815
Mar 24 at 16:44
$begingroup$
@Max0815 if a number is positive, all real exponents of that number will he positive too. Technically, he should have put it around the exponent, but it's really not necessary and in some cases may cause extra confusion. Thats what I think at least
$endgroup$
– Aniruddh Venkatesan
Mar 24 at 17:13
$begingroup$
If $a$ and $b$ are real numbers such that $a^b$ makes sense, then $leftlvert a^brightrvert=lvert arvert^b$.
$endgroup$
– alex.jordan
Mar 24 at 23:25
add a comment |
$begingroup$
In your second case, after you render
$x^fracx^2-5x+42x=-1$
forget about logarithms. Taking absolute values of the equation $a^b=-1$ and remembering that anything to a zero power can only be $+1$, conclude that the only way to get a value of $-1$ using real variables is to have $a=x=-1$. So put $x=-1$ into the equation you derived and see whether, in fact, it's consistent with the power having a value of $-1$.
answered Mar 24 at 9:51
Oscar LanziOscar Lanzi
13.6k12136
$endgroup$
$begingroup$
I see, because no number to any power except (-1) to an odd power gives a value of -1... +1
$endgroup$
– Max0815
Mar 24 at 16:56
add a comment |
$begingroup$
I start off by cross multiplying.
$$x^2=x^x-3+frac4x$$
Taking the square root of both sides gives me:
$$x=pm x^frac12x-frac32+frac42x$$
I start with the positive side first:
$$1=frac12x-frac32+frac42x$$$$x^2-5x+4=0$$$$x=boxed4, boxed1$$
Now, I start with the negative side:
$$x=-x^frac12x-frac32+frac42x$$$$x^fracx^2-5x+42x=-1$$
Absolute value-ing both sides gives $$lvert x^fracx^2-5x+42xrvert=lvert -1rvert$$
Now, we use the property that $lvert x^yrvert=lvert xrvert^y$ to get $$lvert xrvert^fracx^2-5x+42x=1$$
Logging both sides gives:
$$operatornameloglvert xrvert^fracx^2-5x+42x=log1$$$$fracx^2-5x+42xcdotoperatornameloglvert xrvert=0$$$$fracoperatornameloglvert xrvertcdot (x^2-5x+4)2x=0$$
Solving for $operatornameloglvert xrvert$(multiplying by $2x$ and dividing by $x^2-5x+4$) gives:$$operatornameloglvert xrvertcdot (x^2-5x+4)=0$$$$operatornameloglvert xrvert=0$$
Now, solving for $x$ gives:
$$10^operatornameloglvert xrvert=10^0$$$$lvert xrvert=1$$$$x=boxedpm1$$
So the three solutions are $x=-1, 1, 4$ which gives $-1+1+4=boxed4$
answered Mar 24 at 16:55
Max0815Max0815
81418
$endgroup$
$begingroup$
Step 2 above is KEY. Does multiplying by $x^-1$ result in the same magnitude with different signs? I think so but I guess you need to emphasis on this if it is a concrete rule.
$endgroup$
– NoChance
Mar 25 at 11:23
$begingroup$
@NoChance thank you for reminding me. I didn’t see that until my dad reminded me.
$endgroup$
– Max0815
Mar 25 at 15:11
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The reasoning here is faulty:
I can't log both sides because of the $-1$, so I raise both sides by a power of $2x$ to get rid of the fraction. Because $2x$ is even, the RHS becomes a $1$.
For one thing, what if $x$ is not an integer, and $2x$ is not even?
Starting from $x^fracx^2-5x+42x=-1$, you could absolute value both sides: $$lvert xrvert^fracx^2-5x+42x=1$$
and then take the logarithm: $$fracx^2-5x+42xlnlvert xrvert=0$$ from which you conclude either $x=4$, $x=1$, or $lvert xrvert=1$. Check that $4$, $1$, and $-1$ satisfy the original equation, and you have a complete solution.
answered Mar 24 at 5:53
alex.jordanalex.jordan
39.7k660122
$endgroup$
$begingroup$
Hey, a quick question. When you absolute value both sides, why is only the $x$ absolute valued in the LHS, shouldn't it be the whole LHS? Could you explain why?
$endgroup$
– Max0815
Mar 24 at 16:44
$begingroup$
@Max0815 if a number is positive, all real exponents of that number will he positive too. Technically, he should have put it around the exponent, but it's really not necessary and in some cases may cause extra confusion. Thats what I think at least
$endgroup$
– Aniruddh Venkatesan
Mar 24 at 17:13
$begingroup$
If $a$ and $b$ are real numbers such that $a^b$ makes sense, then $leftlvert a^brightrvert=lvert arvert^b$.
$endgroup$
– alex.jordan
Mar 24 at 23:25
add a comment |
$begingroup$
The reasoning here is faulty:
I can't log both sides because of the $-1$, so I raise both sides by a power of $2x$ to get rid of the fraction. Because $2x$ is even, the RHS becomes a $1$.
For one thing, what if $x$ is not an integer, and $2x$ is not even?
Starting from $x^fracx^2-5x+42x=-1$, you could absolute value both sides: $$lvert xrvert^fracx^2-5x+42x=1$$
and then take the logarithm: $$fracx^2-5x+42xlnlvert xrvert=0$$ from which you conclude either $x=4$, $x=1$, or $lvert xrvert=1$. Check that $4$, $1$, and $-1$ satisfy the original equation, and you have a complete solution.
answered Mar 24 at 5:53
alex.jordanalex.jordan
39.7k660122
$endgroup$
$begingroup$
Hey, a quick question. When you absolute value both sides, why is only the $x$ absolute valued in the LHS, shouldn't it be the whole LHS? Could you explain why?
$endgroup$
– Max0815
Mar 24 at 16:44
$begingroup$
@Max0815 if a number is positive, all real exponents of that number will he positive too. Technically, he should have put it around the exponent, but it's really not necessary and in some cases may cause extra confusion. Thats what I think at least
$endgroup$
– Aniruddh Venkatesan
Mar 24 at 17:13
$begingroup$
If $a$ and $b$ are real numbers such that $a^b$ makes sense, then $leftlvert a^brightrvert=lvert arvert^b$.
$endgroup$
– alex.jordan
Mar 24 at 23:25
add a comment |
$begingroup$
The reasoning here is faulty:
I can't log both sides because of the $-1$, so I raise both sides by a power of $2x$ to get rid of the fraction. Because $2x$ is even, the RHS becomes a $1$.
For one thing, what if $x$ is not an integer, and $2x$ is not even?
Starting from $x^fracx^2-5x+42x=-1$, you could absolute value both sides: $$lvert xrvert^fracx^2-5x+42x=1$$
and then take the logarithm: $$fracx^2-5x+42xlnlvert xrvert=0$$ from which you conclude either $x=4$, $x=1$, or $lvert xrvert=1$. Check that $4$, $1$, and $-1$ satisfy the original equation, and you have a complete solution.
answered Mar 24 at 5:53
alex.jordanalex.jordan
39.7k660122
$endgroup$
The reasoning here is faulty:
I can't log both sides because of the $-1$, so I raise both sides by a power of $2x$ to get rid of the fraction. Because $2x$ is even, the RHS becomes a $1$.
For one thing, what if $x$ is not an integer, and $2x$ is not even?
Starting from $x^fracx^2-5x+42x=-1$, you could absolute value both sides: $$lvert xrvert^fracx^2-5x+42x=1$$
and then take the logarithm: $$fracx^2-5x+42xlnlvert xrvert=0$$ from which you conclude either $x=4$, $x=1$, or $lvert xrvert=1$. Check that $4$, $1$, and $-1$ satisfy the original equation, and you have a complete solution.
answered Mar 24 at 5:53
alex.jordanalex.jordan
39.7k660122
answered Mar 24 at 5:53
alex.jordanalex.jordan
39.7k660122
answered Mar 24 at 5:53
alex.jordanalex.jordan
39.7k660122
answered Mar 24 at 5:53
alex.jordanalex.jordan
39.7k660122
39.7k660122
$begingroup$
Hey, a quick question. When you absolute value both sides, why is only the $x$ absolute valued in the LHS, shouldn't it be the whole LHS? Could you explain why?
$endgroup$
– Max0815
Mar 24 at 16:44
$begingroup$
@Max0815 if a number is positive, all real exponents of that number will he positive too. Technically, he should have put it around the exponent, but it's really not necessary and in some cases may cause extra confusion. Thats what I think at least
$endgroup$
– Aniruddh Venkatesan
Mar 24 at 17:13
$begingroup$
If $a$ and $b$ are real numbers such that $a^b$ makes sense, then $leftlvert a^brightrvert=lvert arvert^b$.
$endgroup$
– alex.jordan
Mar 24 at 23:25
add a comment |
$begingroup$
Hey, a quick question. When you absolute value both sides, why is only the $x$ absolute valued in the LHS, shouldn't it be the whole LHS? Could you explain why?
$endgroup$
– Max0815
Mar 24 at 16:44
$begingroup$
@Max0815 if a number is positive, all real exponents of that number will he positive too. Technically, he should have put it around the exponent, but it's really not necessary and in some cases may cause extra confusion. Thats what I think at least
$endgroup$
– Aniruddh Venkatesan
Mar 24 at 17:13
$begingroup$
If $a$ and $b$ are real numbers such that $a^b$ makes sense, then $leftlvert a^brightrvert=lvert arvert^b$.
$endgroup$
– alex.jordan
Mar 24 at 23:25
$begingroup$
Hey, a quick question. When you absolute value both sides, why is only the $x$ absolute valued in the LHS, shouldn't it be the whole LHS? Could you explain why?
$endgroup$
– Max0815
Mar 24 at 16:44
$begingroup$
Hey, a quick question. When you absolute value both sides, why is only the $x$ absolute valued in the LHS, shouldn't it be the whole LHS? Could you explain why?
$endgroup$
– Max0815
Mar 24 at 16:44
$begingroup$
@Max0815 if a number is positive, all real exponents of that number will he positive too. Technically, he should have put it around the exponent, but it's really not necessary and in some cases may cause extra confusion. Thats what I think at least
$endgroup$
– Aniruddh Venkatesan
Mar 24 at 17:13
$begingroup$
@Max0815 if a number is positive, all real exponents of that number will he positive too. Technically, he should have put it around the exponent, but it's really not necessary and in some cases may cause extra confusion. Thats what I think at least
$endgroup$
– Aniruddh Venkatesan
Mar 24 at 17:13
$begingroup$
If $a$ and $b$ are real numbers such that $a^b$ makes sense, then $leftlvert a^brightrvert=lvert arvert^b$.
$endgroup$
– alex.jordan
Mar 24 at 23:25
$begingroup$
If $a$ and $b$ are real numbers such that $a^b$ makes sense, then $leftlvert a^brightrvert=lvert arvert^b$.
$endgroup$
– alex.jordan
Mar 24 at 23:25
add a comment |
$begingroup$
In your second case, after you render
$x^fracx^2-5x+42x=-1$
forget about logarithms. Taking absolute values of the equation $a^b=-1$ and remembering that anything to a zero power can only be $+1$, conclude that the only way to get a value of $-1$ using real variables is to have $a=x=-1$. So put $x=-1$ into the equation you derived and see whether, in fact, it's consistent with the power having a value of $-1$.
answered Mar 24 at 9:51
Oscar LanziOscar Lanzi
13.6k12136
$endgroup$
$begingroup$
I see, because no number to any power except (-1) to an odd power gives a value of -1... +1
$endgroup$
– Max0815
Mar 24 at 16:56
add a comment |
$begingroup$
In your second case, after you render
$x^fracx^2-5x+42x=-1$
forget about logarithms. Taking absolute values of the equation $a^b=-1$ and remembering that anything to a zero power can only be $+1$, conclude that the only way to get a value of $-1$ using real variables is to have $a=x=-1$. So put $x=-1$ into the equation you derived and see whether, in fact, it's consistent with the power having a value of $-1$.
answered Mar 24 at 9:51
Oscar LanziOscar Lanzi
13.6k12136
$endgroup$
$begingroup$
I see, because no number to any power except (-1) to an odd power gives a value of -1... +1
$endgroup$
– Max0815
Mar 24 at 16:56
add a comment |
$begingroup$
In your second case, after you render
$x^fracx^2-5x+42x=-1$
forget about logarithms. Taking absolute values of the equation $a^b=-1$ and remembering that anything to a zero power can only be $+1$, conclude that the only way to get a value of $-1$ using real variables is to have $a=x=-1$. So put $x=-1$ into the equation you derived and see whether, in fact, it's consistent with the power having a value of $-1$.
answered Mar 24 at 9:51
Oscar LanziOscar Lanzi
13.6k12136
$endgroup$
In your second case, after you render
$x^fracx^2-5x+42x=-1$
forget about logarithms. Taking absolute values of the equation $a^b=-1$ and remembering that anything to a zero power can only be $+1$, conclude that the only way to get a value of $-1$ using real variables is to have $a=x=-1$. So put $x=-1$ into the equation you derived and see whether, in fact, it's consistent with the power having a value of $-1$.
answered Mar 24 at 9:51
Oscar LanziOscar Lanzi
13.6k12136
answered Mar 24 at 9:51
Oscar LanziOscar Lanzi
13.6k12136
answered Mar 24 at 9:51
Oscar LanziOscar Lanzi
13.6k12136
answered Mar 24 at 9:51
Oscar LanziOscar Lanzi
13.6k12136
13.6k12136
$begingroup$
I see, because no number to any power except (-1) to an odd power gives a value of -1... +1
$endgroup$
– Max0815
Mar 24 at 16:56
add a comment |
$begingroup$
I see, because no number to any power except (-1) to an odd power gives a value of -1... +1
$endgroup$
– Max0815
Mar 24 at 16:56
$begingroup$
I see, because no number to any power except (-1) to an odd power gives a value of -1... +1
$endgroup$
– Max0815
Mar 24 at 16:56
$begingroup$
I see, because no number to any power except (-1) to an odd power gives a value of -1... +1
$endgroup$
– Max0815
Mar 24 at 16:56
add a comment |
$begingroup$
I start off by cross multiplying.
$$x^2=x^x-3+frac4x$$
Taking the square root of both sides gives me:
$$x=pm x^frac12x-frac32+frac42x$$
I start with the positive side first:
$$1=frac12x-frac32+frac42x$$$$x^2-5x+4=0$$$$x=boxed4, boxed1$$
Now, I start with the negative side:
$$x=-x^frac12x-frac32+frac42x$$$$x^fracx^2-5x+42x=-1$$
Absolute value-ing both sides gives $$lvert x^fracx^2-5x+42xrvert=lvert -1rvert$$
Now, we use the property that $lvert x^yrvert=lvert xrvert^y$ to get $$lvert xrvert^fracx^2-5x+42x=1$$
Logging both sides gives:
$$operatornameloglvert xrvert^fracx^2-5x+42x=log1$$$$fracx^2-5x+42xcdotoperatornameloglvert xrvert=0$$$$fracoperatornameloglvert xrvertcdot (x^2-5x+4)2x=0$$
Solving for $operatornameloglvert xrvert$(multiplying by $2x$ and dividing by $x^2-5x+4$) gives:$$operatornameloglvert xrvertcdot (x^2-5x+4)=0$$$$operatornameloglvert xrvert=0$$
Now, solving for $x$ gives:
$$10^operatornameloglvert xrvert=10^0$$$$lvert xrvert=1$$$$x=boxedpm1$$
So the three solutions are $x=-1, 1, 4$ which gives $-1+1+4=boxed4$
answered Mar 24 at 16:55
Max0815Max0815
81418
$endgroup$
$begingroup$
Step 2 above is KEY. Does multiplying by $x^-1$ result in the same magnitude with different signs? I think so but I guess you need to emphasis on this if it is a concrete rule.
$endgroup$
– NoChance
Mar 25 at 11:23
$begingroup$
@NoChance thank you for reminding me. I didn’t see that until my dad reminded me.
$endgroup$
– Max0815
Mar 25 at 15:11
add a comment |
$begingroup$
I start off by cross multiplying.
$$x^2=x^x-3+frac4x$$
Taking the square root of both sides gives me:
$$x=pm x^frac12x-frac32+frac42x$$
I start with the positive side first:
$$1=frac12x-frac32+frac42x$$$$x^2-5x+4=0$$$$x=boxed4, boxed1$$
Now, I start with the negative side:
$$x=-x^frac12x-frac32+frac42x$$$$x^fracx^2-5x+42x=-1$$
Absolute value-ing both sides gives $$lvert x^fracx^2-5x+42xrvert=lvert -1rvert$$
Now, we use the property that $lvert x^yrvert=lvert xrvert^y$ to get $$lvert xrvert^fracx^2-5x+42x=1$$
Logging both sides gives:
$$operatornameloglvert xrvert^fracx^2-5x+42x=log1$$$$fracx^2-5x+42xcdotoperatornameloglvert xrvert=0$$$$fracoperatornameloglvert xrvertcdot (x^2-5x+4)2x=0$$
Solving for $operatornameloglvert xrvert$(multiplying by $2x$ and dividing by $x^2-5x+4$) gives:$$operatornameloglvert xrvertcdot (x^2-5x+4)=0$$$$operatornameloglvert xrvert=0$$
Now, solving for $x$ gives:
$$10^operatornameloglvert xrvert=10^0$$$$lvert xrvert=1$$$$x=boxedpm1$$
So the three solutions are $x=-1, 1, 4$ which gives $-1+1+4=boxed4$
answered Mar 24 at 16:55
Max0815Max0815
81418
$endgroup$
$begingroup$
Step 2 above is KEY. Does multiplying by $x^-1$ result in the same magnitude with different signs? I think so but I guess you need to emphasis on this if it is a concrete rule.
$endgroup$
– NoChance
Mar 25 at 11:23
$begingroup$
@NoChance thank you for reminding me. I didn’t see that until my dad reminded me.
$endgroup$
– Max0815
Mar 25 at 15:11
add a comment |
$begingroup$
I start off by cross multiplying.
$$x^2=x^x-3+frac4x$$
Taking the square root of both sides gives me:
$$x=pm x^frac12x-frac32+frac42x$$
I start with the positive side first:
$$1=frac12x-frac32+frac42x$$$$x^2-5x+4=0$$$$x=boxed4, boxed1$$
Now, I start with the negative side:
$$x=-x^frac12x-frac32+frac42x$$$$x^fracx^2-5x+42x=-1$$
Absolute value-ing both sides gives $$lvert x^fracx^2-5x+42xrvert=lvert -1rvert$$
Now, we use the property that $lvert x^yrvert=lvert xrvert^y$ to get $$lvert xrvert^fracx^2-5x+42x=1$$
Logging both sides gives:
$$operatornameloglvert xrvert^fracx^2-5x+42x=log1$$$$fracx^2-5x+42xcdotoperatornameloglvert xrvert=0$$$$fracoperatornameloglvert xrvertcdot (x^2-5x+4)2x=0$$
Solving for $operatornameloglvert xrvert$(multiplying by $2x$ and dividing by $x^2-5x+4$) gives:$$operatornameloglvert xrvertcdot (x^2-5x+4)=0$$$$operatornameloglvert xrvert=0$$
Now, solving for $x$ gives:
$$10^operatornameloglvert xrvert=10^0$$$$lvert xrvert=1$$$$x=boxedpm1$$
So the three solutions are $x=-1, 1, 4$ which gives $-1+1+4=boxed4$
answered Mar 24 at 16:55
Max0815Max0815
81418
$endgroup$
I start off by cross multiplying.
$$x^2=x^x-3+frac4x$$
Taking the square root of both sides gives me:
$$x=pm x^frac12x-frac32+frac42x$$
I start with the positive side first:
$$1=frac12x-frac32+frac42x$$$$x^2-5x+4=0$$$$x=boxed4, boxed1$$
Now, I start with the negative side:
$$x=-x^frac12x-frac32+frac42x$$$$x^fracx^2-5x+42x=-1$$
Absolute value-ing both sides gives $$lvert x^fracx^2-5x+42xrvert=lvert -1rvert$$
Now, we use the property that $lvert x^yrvert=lvert xrvert^y$ to get $$lvert xrvert^fracx^2-5x+42x=1$$
Logging both sides gives:
$$operatornameloglvert xrvert^fracx^2-5x+42x=log1$$$$fracx^2-5x+42xcdotoperatornameloglvert xrvert=0$$$$fracoperatornameloglvert xrvertcdot (x^2-5x+4)2x=0$$
Solving for $operatornameloglvert xrvert$(multiplying by $2x$ and dividing by $x^2-5x+4$) gives:$$operatornameloglvert xrvertcdot (x^2-5x+4)=0$$$$operatornameloglvert xrvert=0$$
Now, solving for $x$ gives:
$$10^operatornameloglvert xrvert=10^0$$$$lvert xrvert=1$$$$x=boxedpm1$$
So the three solutions are $x=-1, 1, 4$ which gives $-1+1+4=boxed4$
answered Mar 24 at 16:55
Max0815Max0815
81418
edited Mar 25 at 15:23
answered Mar 24 at 16:55
Max0815Max0815
81418
answered Mar 24 at 16:55
Max0815Max0815
81418
answered Mar 24 at 16:55
Max0815Max0815
81418
81418
$begingroup$
Step 2 above is KEY. Does multiplying by $x^-1$ result in the same magnitude with different signs? I think so but I guess you need to emphasis on this if it is a concrete rule.
$endgroup$
– NoChance
Mar 25 at 11:23
$begingroup$
@NoChance thank you for reminding me. I didn’t see that until my dad reminded me.
$endgroup$
– Max0815
Mar 25 at 15:11
add a comment |
$begingroup$
Step 2 above is KEY. Does multiplying by $x^-1$ result in the same magnitude with different signs? I think so but I guess you need to emphasis on this if it is a concrete rule.
$endgroup$
– NoChance
Mar 25 at 11:23
$begingroup$
@NoChance thank you for reminding me. I didn’t see that until my dad reminded me.
$endgroup$
– Max0815
Mar 25 at 15:11
$begingroup$
Step 2 above is KEY. Does multiplying by $x^-1$ result in the same magnitude with different signs? I think so but I guess you need to emphasis on this if it is a concrete rule.
$endgroup$
– NoChance
Mar 25 at 11:23
$begingroup$
Step 2 above is KEY. Does multiplying by $x^-1$ result in the same magnitude with different signs? I think so but I guess you need to emphasis on this if it is a concrete rule.
$endgroup$
– NoChance
Mar 25 at 11:23
$begingroup$
@NoChance thank you for reminding me. I didn’t see that until my dad reminded me.
$endgroup$
– Max0815
Mar 25 at 15:11
$begingroup$
@NoChance thank you for reminding me. I didn’t see that until my dad reminded me.
$endgroup$
– Max0815
Mar 25 at 15:11
add a comment |
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$begingroup$
See math.stackexchange.com/questions/3157637/…
$endgroup$
– lab bhattacharjee
Mar 24 at 6:13
$begingroup$
This is a good question. In fact if you draw the 2 curves, x=-1 does not show as an intersection point !!!
$endgroup$
– NoChance
Mar 24 at 7:22
$begingroup$
Technically it does. While continuous curves can be rendered only in the first quadrant, the functions over $mathbbR$ also include other individual points where they may be defined. One such point for both functions is $x=-1$, and at that value of $x$ the points are the same for the two functions.
$endgroup$
– Oscar Lanzi
Mar 24 at 11:18
$begingroup$
@OscarLanzi, thank your for your comment. I am still not clear why would a program not plot the said curves at x=-1 if the function is defined at x=-1?
$endgroup$
– NoChance
Mar 24 at 13:48
$begingroup$
@OscarLanzi, thank your for your comment. I was not aware of the relationship between zeroes, algebraic operations and continuity of the function....
$endgroup$
– NoChance
Mar 24 at 14:27