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Why am I allowed to remove $ln$ from both sides of an equation?
Solving an equation when the unknown is both a term and exponentWhat operations on an equation cause it to be destroyed?Deriving a power equation from a log-log line equationWhy do I receive the wrong answer when I try to solve this exponential equation?Why can we take the log of both sides?Solving the equation $sqrt[3]x^2 + 15 = 2sqrt[3]x+1$What is intresting about $sqrtlog_xexpsqrtlog_xexpsqrtlog_xexpcdots=log_xe$?Taking the same thing of both sides of an equationWolfram Alpha gives different answers for doing the same thing to both sides of this equation.Why does taking logarithms on both sides of $0<r<s$ reverse the inequality for logarithms with base $a$, $0<a<1$?
$begingroup$
When I have this equation: $ln(x+2) = ln(x^2)$, why can I just remove the $ln$ from both sides by raising it to the power of e. Does this not permanently change what the equitation equals? Also, when I graph $x+2=x^2$ and $ln(x+2) = ln(x^2)$ the roots of the equation are the same. Why are they?
Q1: Why can I remove the $ln$ by raising both sides of the equation to e? Does this not permanently change what the equation is?
Q2: Why are the roots of the two equations above the same?
logarithms roots
$endgroup$
add a comment |
$begingroup$
When I have this equation: $ln(x+2) = ln(x^2)$, why can I just remove the $ln$ from both sides by raising it to the power of e. Does this not permanently change what the equitation equals? Also, when I graph $x+2=x^2$ and $ln(x+2) = ln(x^2)$ the roots of the equation are the same. Why are they?
Q1: Why can I remove the $ln$ by raising both sides of the equation to e? Does this not permanently change what the equation is?
Q2: Why are the roots of the two equations above the same?
logarithms roots
$endgroup$
4
$begingroup$
logarithms (and many other functions) have a property called "injectivity" which makes the trick. A function which is not injective is not able to perform such a magic. The classical example of a function which is not injective is $x^2$. If $x^2 = 2^2$ you can't remove the squares, because you will lose the negative solution.
$endgroup$
– FormerMath
Mar 17 at 23:27
$begingroup$
You should include in your problem statement what kind of number $x$ is known to be when the first equation is given.
$endgroup$
– hardmath
Mar 17 at 23:27
add a comment |
$begingroup$
When I have this equation: $ln(x+2) = ln(x^2)$, why can I just remove the $ln$ from both sides by raising it to the power of e. Does this not permanently change what the equitation equals? Also, when I graph $x+2=x^2$ and $ln(x+2) = ln(x^2)$ the roots of the equation are the same. Why are they?
Q1: Why can I remove the $ln$ by raising both sides of the equation to e? Does this not permanently change what the equation is?
Q2: Why are the roots of the two equations above the same?
logarithms roots
$endgroup$
When I have this equation: $ln(x+2) = ln(x^2)$, why can I just remove the $ln$ from both sides by raising it to the power of e. Does this not permanently change what the equitation equals? Also, when I graph $x+2=x^2$ and $ln(x+2) = ln(x^2)$ the roots of the equation are the same. Why are they?
Q1: Why can I remove the $ln$ by raising both sides of the equation to e? Does this not permanently change what the equation is?
Q2: Why are the roots of the two equations above the same?
logarithms roots
logarithms roots
asked Mar 17 at 23:21
Rory ShaughnessyRory Shaughnessy
132
132
4
$begingroup$
logarithms (and many other functions) have a property called "injectivity" which makes the trick. A function which is not injective is not able to perform such a magic. The classical example of a function which is not injective is $x^2$. If $x^2 = 2^2$ you can't remove the squares, because you will lose the negative solution.
$endgroup$
– FormerMath
Mar 17 at 23:27
$begingroup$
You should include in your problem statement what kind of number $x$ is known to be when the first equation is given.
$endgroup$
– hardmath
Mar 17 at 23:27
add a comment |
4
$begingroup$
logarithms (and many other functions) have a property called "injectivity" which makes the trick. A function which is not injective is not able to perform such a magic. The classical example of a function which is not injective is $x^2$. If $x^2 = 2^2$ you can't remove the squares, because you will lose the negative solution.
$endgroup$
– FormerMath
Mar 17 at 23:27
$begingroup$
You should include in your problem statement what kind of number $x$ is known to be when the first equation is given.
$endgroup$
– hardmath
Mar 17 at 23:27
4
4
$begingroup$
logarithms (and many other functions) have a property called "injectivity" which makes the trick. A function which is not injective is not able to perform such a magic. The classical example of a function which is not injective is $x^2$. If $x^2 = 2^2$ you can't remove the squares, because you will lose the negative solution.
$endgroup$
– FormerMath
Mar 17 at 23:27
$begingroup$
logarithms (and many other functions) have a property called "injectivity" which makes the trick. A function which is not injective is not able to perform such a magic. The classical example of a function which is not injective is $x^2$. If $x^2 = 2^2$ you can't remove the squares, because you will lose the negative solution.
$endgroup$
– FormerMath
Mar 17 at 23:27
$begingroup$
You should include in your problem statement what kind of number $x$ is known to be when the first equation is given.
$endgroup$
– hardmath
Mar 17 at 23:27
$begingroup$
You should include in your problem statement what kind of number $x$ is known to be when the first equation is given.
$endgroup$
– hardmath
Mar 17 at 23:27
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
Equations don't equal anything. Equations are sentences. You can replace a sentence with a different sentence as long as it is also true. 1)$27x = 54y$ is a sentence. 2)$x = 2y$ is a different sentence. The first sentence in only true if the second one is too and vice versa. But they are different sentences. We can replace one with the other because one follows from the other. We don't care what $27x$ equals. We care what $x$ equals.
If $a = b$ then $anythinginvolving(a) = anythinginvolving(b)$. So if $a = b$ then $e^a = e^b$. Likewise $a + 27 = b+ 27$ or $a^2 - sqrt3a + 5 = b^2 -sqrt3b + 5$. If $a = b$ we can do anything to either side.
So if $ln (x+2) = ln x^2$ we can do $e^ln(x+2) = e^ln x^2$. Why not? If we wanted to, we could say $ln(x+2)$ fried in batter and stuffed inside $sqrttexta tomato=ln(x^2)$ fried in batter and stuffed inside $sqrttexta tomato$ for all anyone cares.
But $e^ln(x+2)= e^ln x^2$ is a smart thing to care about. (Whereas $ln(x+2)$ fried in batter and stuffed inside $sqrttexta tomato=ln(x^2)$ fried in batter and stuffed inside $sqrttexta tomato$ is an utterly stupid thing to care about.)
BY DEFINITION: $e^ln a = a$, ALWAYS. And for real numbers $ln e^a = a$. $e^x$ and $ln x$ are inverses and that means they can "undo" each other. Ith this way they are just like multiplication and division. If you have $3(x+2) = 3x^2$ you can "undo" the "times $3$ by dividing both sides by $3$ because division "undoes" multiplication.
So $ln(x+2) = ln x^2$ is a true sentence. We don't care.
That means $e^ln(x+2) = e^ln x^2$ is a different true sentence.
But this different true sentence means the same thing as:
$x+2 = e^ln(x+1) = e^ln x^2 = x^2$ so now we have yet another true sentence:
$x + 2= x^2$. And we care because the takes us one step closer to soling what $x$ is.
====
Postscipt:
If $a = b$ we can do anything to either side and get a true statement.
BUt be careful. Not everything we do can be undone.
For example: If $x = 7$ then $3x + 5 = 3*7 + 5 = 26$. And $0times (3x+5) = 0times 26$ and so $0 = 0$. These are all true sentences. But the are not equivalent sentences. Multiplying by $0$ can not be "undone". But multiplying by $3$ can be undone (by dividing by $3$) and adding $5$ can be undone (by subtracting $5$).
Raising $e$ to a power can be undone by taking the logorithm. And taking the logorithm can be undone by raising $e$ to the power but only if the number was positive to begin with.
$endgroup$
add a comment |
$begingroup$
Q1: We have $$ln g(x) = ln f(x) implies ln g(x) - ln f(x) = 0 implies ln dfracg(x)f(x)=0 implies dfracg(x)f(x)=1.$$
Q2: Note that the previous operations are only valid if $ln g(x)$ and $ln f(x)$ are defined. We get into trouble if the arguments are allowed to be negative or zero. The roots are the same because the solution to the equation happens to be in the allowed region.
$endgroup$
add a comment |
$begingroup$
Q1. Because of injectivity of the log.
Q2. Same answer as Q1
$endgroup$
add a comment |
$begingroup$
The mapping
beginarraycccc
ln: &(0,infty) &to & mathbb R \
& x &mapsto &ln(x)
endarray
is one-to-one onto, and strictly increasing.
Similarly, the mapping
beginarraycccc
exp: & mathbb R &to &(0,infty) \
& x &mapsto &exp(x)=e^x
endarray
is one-to-one onto, and strictly increasing.
- $x < y iff e^x < e^y$
- $x = y iff e^x = e^y$
- $x < y iff ln(x) < ln(y)$
- $x = y iff ln(x) = ln(y)$
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
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active
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$begingroup$
Equations don't equal anything. Equations are sentences. You can replace a sentence with a different sentence as long as it is also true. 1)$27x = 54y$ is a sentence. 2)$x = 2y$ is a different sentence. The first sentence in only true if the second one is too and vice versa. But they are different sentences. We can replace one with the other because one follows from the other. We don't care what $27x$ equals. We care what $x$ equals.
If $a = b$ then $anythinginvolving(a) = anythinginvolving(b)$. So if $a = b$ then $e^a = e^b$. Likewise $a + 27 = b+ 27$ or $a^2 - sqrt3a + 5 = b^2 -sqrt3b + 5$. If $a = b$ we can do anything to either side.
So if $ln (x+2) = ln x^2$ we can do $e^ln(x+2) = e^ln x^2$. Why not? If we wanted to, we could say $ln(x+2)$ fried in batter and stuffed inside $sqrttexta tomato=ln(x^2)$ fried in batter and stuffed inside $sqrttexta tomato$ for all anyone cares.
But $e^ln(x+2)= e^ln x^2$ is a smart thing to care about. (Whereas $ln(x+2)$ fried in batter and stuffed inside $sqrttexta tomato=ln(x^2)$ fried in batter and stuffed inside $sqrttexta tomato$ is an utterly stupid thing to care about.)
BY DEFINITION: $e^ln a = a$, ALWAYS. And for real numbers $ln e^a = a$. $e^x$ and $ln x$ are inverses and that means they can "undo" each other. Ith this way they are just like multiplication and division. If you have $3(x+2) = 3x^2$ you can "undo" the "times $3$ by dividing both sides by $3$ because division "undoes" multiplication.
So $ln(x+2) = ln x^2$ is a true sentence. We don't care.
That means $e^ln(x+2) = e^ln x^2$ is a different true sentence.
But this different true sentence means the same thing as:
$x+2 = e^ln(x+1) = e^ln x^2 = x^2$ so now we have yet another true sentence:
$x + 2= x^2$. And we care because the takes us one step closer to soling what $x$ is.
====
Postscipt:
If $a = b$ we can do anything to either side and get a true statement.
BUt be careful. Not everything we do can be undone.
For example: If $x = 7$ then $3x + 5 = 3*7 + 5 = 26$. And $0times (3x+5) = 0times 26$ and so $0 = 0$. These are all true sentences. But the are not equivalent sentences. Multiplying by $0$ can not be "undone". But multiplying by $3$ can be undone (by dividing by $3$) and adding $5$ can be undone (by subtracting $5$).
Raising $e$ to a power can be undone by taking the logorithm. And taking the logorithm can be undone by raising $e$ to the power but only if the number was positive to begin with.
$endgroup$
add a comment |
$begingroup$
Equations don't equal anything. Equations are sentences. You can replace a sentence with a different sentence as long as it is also true. 1)$27x = 54y$ is a sentence. 2)$x = 2y$ is a different sentence. The first sentence in only true if the second one is too and vice versa. But they are different sentences. We can replace one with the other because one follows from the other. We don't care what $27x$ equals. We care what $x$ equals.
If $a = b$ then $anythinginvolving(a) = anythinginvolving(b)$. So if $a = b$ then $e^a = e^b$. Likewise $a + 27 = b+ 27$ or $a^2 - sqrt3a + 5 = b^2 -sqrt3b + 5$. If $a = b$ we can do anything to either side.
So if $ln (x+2) = ln x^2$ we can do $e^ln(x+2) = e^ln x^2$. Why not? If we wanted to, we could say $ln(x+2)$ fried in batter and stuffed inside $sqrttexta tomato=ln(x^2)$ fried in batter and stuffed inside $sqrttexta tomato$ for all anyone cares.
But $e^ln(x+2)= e^ln x^2$ is a smart thing to care about. (Whereas $ln(x+2)$ fried in batter and stuffed inside $sqrttexta tomato=ln(x^2)$ fried in batter and stuffed inside $sqrttexta tomato$ is an utterly stupid thing to care about.)
BY DEFINITION: $e^ln a = a$, ALWAYS. And for real numbers $ln e^a = a$. $e^x$ and $ln x$ are inverses and that means they can "undo" each other. Ith this way they are just like multiplication and division. If you have $3(x+2) = 3x^2$ you can "undo" the "times $3$ by dividing both sides by $3$ because division "undoes" multiplication.
So $ln(x+2) = ln x^2$ is a true sentence. We don't care.
That means $e^ln(x+2) = e^ln x^2$ is a different true sentence.
But this different true sentence means the same thing as:
$x+2 = e^ln(x+1) = e^ln x^2 = x^2$ so now we have yet another true sentence:
$x + 2= x^2$. And we care because the takes us one step closer to soling what $x$ is.
====
Postscipt:
If $a = b$ we can do anything to either side and get a true statement.
BUt be careful. Not everything we do can be undone.
For example: If $x = 7$ then $3x + 5 = 3*7 + 5 = 26$. And $0times (3x+5) = 0times 26$ and so $0 = 0$. These are all true sentences. But the are not equivalent sentences. Multiplying by $0$ can not be "undone". But multiplying by $3$ can be undone (by dividing by $3$) and adding $5$ can be undone (by subtracting $5$).
Raising $e$ to a power can be undone by taking the logorithm. And taking the logorithm can be undone by raising $e$ to the power but only if the number was positive to begin with.
$endgroup$
add a comment |
$begingroup$
Equations don't equal anything. Equations are sentences. You can replace a sentence with a different sentence as long as it is also true. 1)$27x = 54y$ is a sentence. 2)$x = 2y$ is a different sentence. The first sentence in only true if the second one is too and vice versa. But they are different sentences. We can replace one with the other because one follows from the other. We don't care what $27x$ equals. We care what $x$ equals.
If $a = b$ then $anythinginvolving(a) = anythinginvolving(b)$. So if $a = b$ then $e^a = e^b$. Likewise $a + 27 = b+ 27$ or $a^2 - sqrt3a + 5 = b^2 -sqrt3b + 5$. If $a = b$ we can do anything to either side.
So if $ln (x+2) = ln x^2$ we can do $e^ln(x+2) = e^ln x^2$. Why not? If we wanted to, we could say $ln(x+2)$ fried in batter and stuffed inside $sqrttexta tomato=ln(x^2)$ fried in batter and stuffed inside $sqrttexta tomato$ for all anyone cares.
But $e^ln(x+2)= e^ln x^2$ is a smart thing to care about. (Whereas $ln(x+2)$ fried in batter and stuffed inside $sqrttexta tomato=ln(x^2)$ fried in batter and stuffed inside $sqrttexta tomato$ is an utterly stupid thing to care about.)
BY DEFINITION: $e^ln a = a$, ALWAYS. And for real numbers $ln e^a = a$. $e^x$ and $ln x$ are inverses and that means they can "undo" each other. Ith this way they are just like multiplication and division. If you have $3(x+2) = 3x^2$ you can "undo" the "times $3$ by dividing both sides by $3$ because division "undoes" multiplication.
So $ln(x+2) = ln x^2$ is a true sentence. We don't care.
That means $e^ln(x+2) = e^ln x^2$ is a different true sentence.
But this different true sentence means the same thing as:
$x+2 = e^ln(x+1) = e^ln x^2 = x^2$ so now we have yet another true sentence:
$x + 2= x^2$. And we care because the takes us one step closer to soling what $x$ is.
====
Postscipt:
If $a = b$ we can do anything to either side and get a true statement.
BUt be careful. Not everything we do can be undone.
For example: If $x = 7$ then $3x + 5 = 3*7 + 5 = 26$. And $0times (3x+5) = 0times 26$ and so $0 = 0$. These are all true sentences. But the are not equivalent sentences. Multiplying by $0$ can not be "undone". But multiplying by $3$ can be undone (by dividing by $3$) and adding $5$ can be undone (by subtracting $5$).
Raising $e$ to a power can be undone by taking the logorithm. And taking the logorithm can be undone by raising $e$ to the power but only if the number was positive to begin with.
$endgroup$
Equations don't equal anything. Equations are sentences. You can replace a sentence with a different sentence as long as it is also true. 1)$27x = 54y$ is a sentence. 2)$x = 2y$ is a different sentence. The first sentence in only true if the second one is too and vice versa. But they are different sentences. We can replace one with the other because one follows from the other. We don't care what $27x$ equals. We care what $x$ equals.
If $a = b$ then $anythinginvolving(a) = anythinginvolving(b)$. So if $a = b$ then $e^a = e^b$. Likewise $a + 27 = b+ 27$ or $a^2 - sqrt3a + 5 = b^2 -sqrt3b + 5$. If $a = b$ we can do anything to either side.
So if $ln (x+2) = ln x^2$ we can do $e^ln(x+2) = e^ln x^2$. Why not? If we wanted to, we could say $ln(x+2)$ fried in batter and stuffed inside $sqrttexta tomato=ln(x^2)$ fried in batter and stuffed inside $sqrttexta tomato$ for all anyone cares.
But $e^ln(x+2)= e^ln x^2$ is a smart thing to care about. (Whereas $ln(x+2)$ fried in batter and stuffed inside $sqrttexta tomato=ln(x^2)$ fried in batter and stuffed inside $sqrttexta tomato$ is an utterly stupid thing to care about.)
BY DEFINITION: $e^ln a = a$, ALWAYS. And for real numbers $ln e^a = a$. $e^x$ and $ln x$ are inverses and that means they can "undo" each other. Ith this way they are just like multiplication and division. If you have $3(x+2) = 3x^2$ you can "undo" the "times $3$ by dividing both sides by $3$ because division "undoes" multiplication.
So $ln(x+2) = ln x^2$ is a true sentence. We don't care.
That means $e^ln(x+2) = e^ln x^2$ is a different true sentence.
But this different true sentence means the same thing as:
$x+2 = e^ln(x+1) = e^ln x^2 = x^2$ so now we have yet another true sentence:
$x + 2= x^2$. And we care because the takes us one step closer to soling what $x$ is.
====
Postscipt:
If $a = b$ we can do anything to either side and get a true statement.
BUt be careful. Not everything we do can be undone.
For example: If $x = 7$ then $3x + 5 = 3*7 + 5 = 26$. And $0times (3x+5) = 0times 26$ and so $0 = 0$. These are all true sentences. But the are not equivalent sentences. Multiplying by $0$ can not be "undone". But multiplying by $3$ can be undone (by dividing by $3$) and adding $5$ can be undone (by subtracting $5$).
Raising $e$ to a power can be undone by taking the logorithm. And taking the logorithm can be undone by raising $e$ to the power but only if the number was positive to begin with.
answered Mar 18 at 0:13
fleabloodfleablood
73.5k22891
73.5k22891
add a comment |
add a comment |
$begingroup$
Q1: We have $$ln g(x) = ln f(x) implies ln g(x) - ln f(x) = 0 implies ln dfracg(x)f(x)=0 implies dfracg(x)f(x)=1.$$
Q2: Note that the previous operations are only valid if $ln g(x)$ and $ln f(x)$ are defined. We get into trouble if the arguments are allowed to be negative or zero. The roots are the same because the solution to the equation happens to be in the allowed region.
$endgroup$
add a comment |
$begingroup$
Q1: We have $$ln g(x) = ln f(x) implies ln g(x) - ln f(x) = 0 implies ln dfracg(x)f(x)=0 implies dfracg(x)f(x)=1.$$
Q2: Note that the previous operations are only valid if $ln g(x)$ and $ln f(x)$ are defined. We get into trouble if the arguments are allowed to be negative or zero. The roots are the same because the solution to the equation happens to be in the allowed region.
$endgroup$
add a comment |
$begingroup$
Q1: We have $$ln g(x) = ln f(x) implies ln g(x) - ln f(x) = 0 implies ln dfracg(x)f(x)=0 implies dfracg(x)f(x)=1.$$
Q2: Note that the previous operations are only valid if $ln g(x)$ and $ln f(x)$ are defined. We get into trouble if the arguments are allowed to be negative or zero. The roots are the same because the solution to the equation happens to be in the allowed region.
$endgroup$
Q1: We have $$ln g(x) = ln f(x) implies ln g(x) - ln f(x) = 0 implies ln dfracg(x)f(x)=0 implies dfracg(x)f(x)=1.$$
Q2: Note that the previous operations are only valid if $ln g(x)$ and $ln f(x)$ are defined. We get into trouble if the arguments are allowed to be negative or zero. The roots are the same because the solution to the equation happens to be in the allowed region.
answered Mar 17 at 23:25
MachineLearnerMachineLearner
1,319112
1,319112
add a comment |
add a comment |
$begingroup$
Q1. Because of injectivity of the log.
Q2. Same answer as Q1
$endgroup$
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Q1. Because of injectivity of the log.
Q2. Same answer as Q1
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Q1. Because of injectivity of the log.
Q2. Same answer as Q1
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Q1. Because of injectivity of the log.
Q2. Same answer as Q1
answered Mar 17 at 23:25
HAMIDINE SOUMAREHAMIDINE SOUMARE
1
1
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The mapping
beginarraycccc
ln: &(0,infty) &to & mathbb R \
& x &mapsto &ln(x)
endarray
is one-to-one onto, and strictly increasing.
Similarly, the mapping
beginarraycccc
exp: & mathbb R &to &(0,infty) \
& x &mapsto &exp(x)=e^x
endarray
is one-to-one onto, and strictly increasing.
- $x < y iff e^x < e^y$
- $x = y iff e^x = e^y$
- $x < y iff ln(x) < ln(y)$
- $x = y iff ln(x) = ln(y)$
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add a comment |
$begingroup$
The mapping
beginarraycccc
ln: &(0,infty) &to & mathbb R \
& x &mapsto &ln(x)
endarray
is one-to-one onto, and strictly increasing.
Similarly, the mapping
beginarraycccc
exp: & mathbb R &to &(0,infty) \
& x &mapsto &exp(x)=e^x
endarray
is one-to-one onto, and strictly increasing.
- $x < y iff e^x < e^y$
- $x = y iff e^x = e^y$
- $x < y iff ln(x) < ln(y)$
- $x = y iff ln(x) = ln(y)$
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add a comment |
$begingroup$
The mapping
beginarraycccc
ln: &(0,infty) &to & mathbb R \
& x &mapsto &ln(x)
endarray
is one-to-one onto, and strictly increasing.
Similarly, the mapping
beginarraycccc
exp: & mathbb R &to &(0,infty) \
& x &mapsto &exp(x)=e^x
endarray
is one-to-one onto, and strictly increasing.
- $x < y iff e^x < e^y$
- $x = y iff e^x = e^y$
- $x < y iff ln(x) < ln(y)$
- $x = y iff ln(x) = ln(y)$
$endgroup$
The mapping
beginarraycccc
ln: &(0,infty) &to & mathbb R \
& x &mapsto &ln(x)
endarray
is one-to-one onto, and strictly increasing.
Similarly, the mapping
beginarraycccc
exp: & mathbb R &to &(0,infty) \
& x &mapsto &exp(x)=e^x
endarray
is one-to-one onto, and strictly increasing.
- $x < y iff e^x < e^y$
- $x = y iff e^x = e^y$
- $x < y iff ln(x) < ln(y)$
- $x = y iff ln(x) = ln(y)$
edited Mar 20 at 17:15
answered Mar 17 at 23:44
steven gregorysteven gregory
18.3k32358
18.3k32358
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logarithms (and many other functions) have a property called "injectivity" which makes the trick. A function which is not injective is not able to perform such a magic. The classical example of a function which is not injective is $x^2$. If $x^2 = 2^2$ you can't remove the squares, because you will lose the negative solution.
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– FormerMath
Mar 17 at 23:27
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You should include in your problem statement what kind of number $x$ is known to be when the first equation is given.
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– hardmath
Mar 17 at 23:27