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What is the answer to this exponential equation?
Simple proof of the Binomial Theorem for $mathbbR$Simplifying the expression of exponential and logarithmshow can I show that $a^nrightarrow 0$ if $0<a<1$Solving an exponential equation without the quadratic formulaHow do i convince students in high school for which this equation: $2^x=4x$ have only one solution in integers that is $x=4$?Approximating the value of a definite integralWhat are some clever (preferably physical) examples of processes exhibiting exponential growth?Solving for $x$ in this equationIs there any simple way to solve $366^n(366-n)!geq 2times 366!$?How many solutions are there for the equation $a^x = log_a x$, where $0 < a < 1$?
$begingroup$
If
$$
3^x-2=7,
$$
what is the result of
$$
(144)^frac12x-2=?
$$
This is a question for high school students. Response options,
$$
textA. mathbf7text B. mathbf49text C. mathbf
3text D. mathbf9text E. mathbffrac13
$$
analysis inequality logarithms exponential-function
asked 15 hours ago
Abdullah AKKURTAbdullah AKKURT
13
New contributor
Abdullah AKKURT is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
|
show 1 more comment
$begingroup$
If
$$
3^x-2=7,
$$
what is the result of
$$
(144)^frac12x-2=?
$$
This is a question for high school students. Response options,
$$
textA. mathbf7text B. mathbf49text C. mathbf
3text D. mathbf9text E. mathbffrac13
$$
analysis inequality logarithms exponential-function
asked 15 hours ago
Abdullah AKKURTAbdullah AKKURT
13
New contributor
Abdullah AKKURT is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
Pretty sure the problem is given or copied wrong.
$endgroup$
– Ian
15 hours ago
$begingroup$
$144$ or $441$??
$endgroup$
– E.H.E
15 hours ago
$begingroup$
None of them. Are you sure about the formula ?
$endgroup$
– Claude Leibovici
15 hours ago
$begingroup$
The question is right. There are no errors.
$endgroup$
– Abdullah AKKURT
15 hours ago
2
$begingroup$
@AbdullahAKKURT It's not right. The result is not one of the options given.
$endgroup$
– Eff
15 hours ago
|
show 1 more comment
$begingroup$
If
$$
3^x-2=7,
$$
what is the result of
$$
(144)^frac12x-2=?
$$
This is a question for high school students. Response options,
$$
textA. mathbf7text B. mathbf49text C. mathbf
3text D. mathbf9text E. mathbffrac13
$$
analysis inequality logarithms exponential-function
asked 15 hours ago
Abdullah AKKURTAbdullah AKKURT
13
New contributor
Abdullah AKKURT is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
If
$$
3^x-2=7,
$$
what is the result of
$$
(144)^frac12x-2=?
$$
This is a question for high school students. Response options,
$$
textA. mathbf7text B. mathbf49text C. mathbf
3text D. mathbf9text E. mathbffrac13
$$
analysis inequality logarithms exponential-function
analysis inequality logarithms exponential-function
asked 15 hours ago
Abdullah AKKURTAbdullah AKKURT
13
New contributor
Abdullah AKKURT is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 15 hours ago
Abdullah AKKURTAbdullah AKKURT
13
New contributor
Abdullah AKKURT is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 14 hours ago
Abdullah AKKURT
asked 15 hours ago
Abdullah AKKURTAbdullah AKKURT
13
New contributor
Abdullah AKKURT is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 15 hours ago
Abdullah AKKURTAbdullah AKKURT
13
asked 15 hours ago
Abdullah AKKURTAbdullah AKKURT
13
13
New contributor
Abdullah AKKURT is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Abdullah AKKURT is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Abdullah AKKURT is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
Pretty sure the problem is given or copied wrong.
$endgroup$
– Ian
15 hours ago
$begingroup$
$144$ or $441$??
$endgroup$
– E.H.E
15 hours ago
$begingroup$
None of them. Are you sure about the formula ?
$endgroup$
– Claude Leibovici
15 hours ago
$begingroup$
The question is right. There are no errors.
$endgroup$
– Abdullah AKKURT
15 hours ago
2
$begingroup$
@AbdullahAKKURT It's not right. The result is not one of the options given.
$endgroup$
– Eff
15 hours ago
|
show 1 more comment
$begingroup$
Pretty sure the problem is given or copied wrong.
$endgroup$
– Ian
15 hours ago
$begingroup$
$144$ or $441$??
$endgroup$
– E.H.E
15 hours ago
$begingroup$
None of them. Are you sure about the formula ?
$endgroup$
– Claude Leibovici
15 hours ago
$begingroup$
The question is right. There are no errors.
$endgroup$
– Abdullah AKKURT
15 hours ago
2
$begingroup$
@AbdullahAKKURT It's not right. The result is not one of the options given.
$endgroup$
– Eff
15 hours ago
$begingroup$
Pretty sure the problem is given or copied wrong.
$endgroup$
– Ian
15 hours ago
$begingroup$
Pretty sure the problem is given or copied wrong.
$endgroup$
– Ian
15 hours ago
$begingroup$
$144$ or $441$??
$endgroup$
– E.H.E
15 hours ago
$begingroup$
$144$ or $441$??
$endgroup$
– E.H.E
15 hours ago
$begingroup$
None of them. Are you sure about the formula ?
$endgroup$
– Claude Leibovici
15 hours ago
$begingroup$
None of them. Are you sure about the formula ?
$endgroup$
– Claude Leibovici
15 hours ago
$begingroup$
The question is right. There are no errors.
$endgroup$
– Abdullah AKKURT
15 hours ago
$begingroup$
The question is right. There are no errors.
$endgroup$
– Abdullah AKKURT
15 hours ago
2
2
$begingroup$
@AbdullahAKKURT It's not right. The result is not one of the options given.
$endgroup$
– Eff
15 hours ago
$begingroup$
@AbdullahAKKURT It's not right. The result is not one of the options given.
$endgroup$
– Eff
15 hours ago
|
show 1 more comment
2 Answers
2
active
oldest
votes
$begingroup$
I think there is a wrong in number $144$ which should be $441$ as according to this answer
$$3^x-2=7$$
$$frac3^x9=7$$
$$3^x=63$$
$$3^2x=3969$$
$$3^2x-2=frac39699$$
$$3^2x-2=441$$
so
$$(441)^frac12x-2=3$$
therefore the answer is $C$
edited 14 hours ago
Toby Mak
3,53111128
answered 15 hours ago
E.H.EE.H.E
15.8k11968
$endgroup$
$begingroup$
A logical solution, however, is given as 144 instead of 441. Otherwise the solution is very easy.
$endgroup$
– Abdullah AKKURT
14 hours ago
1
$begingroup$
@AbdullahAKKURT But with $144$ the answer is an irrational number. If the solution is that straightforward, there is no need to complicate things.
$endgroup$
– Toby Mak
14 hours ago
add a comment |
$begingroup$
My solution
begineqnarray*
3^1 &<&3^x-2<3^2 \
1 &<&x-2<2 \
3 &<&x<4 \
4 &<&2x-2<6 \
frac16 &<½x-2<frac14.
endeqnarray*
Thus,
begineqnarray*
144^frac16 &<&144^frac12x-2<144^frac14 \
2.28943... &<&144^frac12x-2<3.46410...
endeqnarray*
The logical answer is $mathbf3$.
answered 14 hours ago
Abdullah AKKURTAbdullah AKKURT
13
New contributor
Abdullah AKKURT is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I think there is a wrong in number $144$ which should be $441$ as according to this answer
$$3^x-2=7$$
$$frac3^x9=7$$
$$3^x=63$$
$$3^2x=3969$$
$$3^2x-2=frac39699$$
$$3^2x-2=441$$
so
$$(441)^frac12x-2=3$$
therefore the answer is $C$
edited 14 hours ago
Toby Mak
3,53111128
answered 15 hours ago
E.H.EE.H.E
15.8k11968
$endgroup$
$begingroup$
A logical solution, however, is given as 144 instead of 441. Otherwise the solution is very easy.
$endgroup$
– Abdullah AKKURT
14 hours ago
1
$begingroup$
@AbdullahAKKURT But with $144$ the answer is an irrational number. If the solution is that straightforward, there is no need to complicate things.
$endgroup$
– Toby Mak
14 hours ago
add a comment |
$begingroup$
I think there is a wrong in number $144$ which should be $441$ as according to this answer
$$3^x-2=7$$
$$frac3^x9=7$$
$$3^x=63$$
$$3^2x=3969$$
$$3^2x-2=frac39699$$
$$3^2x-2=441$$
so
$$(441)^frac12x-2=3$$
therefore the answer is $C$
edited 14 hours ago
Toby Mak
3,53111128
answered 15 hours ago
E.H.EE.H.E
15.8k11968
$endgroup$
$begingroup$
A logical solution, however, is given as 144 instead of 441. Otherwise the solution is very easy.
$endgroup$
– Abdullah AKKURT
14 hours ago
1
$begingroup$
@AbdullahAKKURT But with $144$ the answer is an irrational number. If the solution is that straightforward, there is no need to complicate things.
$endgroup$
– Toby Mak
14 hours ago
add a comment |
$begingroup$
I think there is a wrong in number $144$ which should be $441$ as according to this answer
$$3^x-2=7$$
$$frac3^x9=7$$
$$3^x=63$$
$$3^2x=3969$$
$$3^2x-2=frac39699$$
$$3^2x-2=441$$
so
$$(441)^frac12x-2=3$$
therefore the answer is $C$
edited 14 hours ago
Toby Mak
3,53111128
answered 15 hours ago
E.H.EE.H.E
15.8k11968
$endgroup$
I think there is a wrong in number $144$ which should be $441$ as according to this answer
$$3^x-2=7$$
$$frac3^x9=7$$
$$3^x=63$$
$$3^2x=3969$$
$$3^2x-2=frac39699$$
$$3^2x-2=441$$
so
$$(441)^frac12x-2=3$$
therefore the answer is $C$
edited 14 hours ago
Toby Mak
3,53111128
answered 15 hours ago
E.H.EE.H.E
15.8k11968
edited 14 hours ago
Toby Mak
3,53111128
edited 14 hours ago
Toby Mak
3,53111128
edited 14 hours ago
Toby Mak
3,53111128
3,53111128
answered 15 hours ago
E.H.EE.H.E
15.8k11968
answered 15 hours ago
E.H.EE.H.E
15.8k11968
answered 15 hours ago
E.H.EE.H.E
15.8k11968
15.8k11968
$begingroup$
A logical solution, however, is given as 144 instead of 441. Otherwise the solution is very easy.
$endgroup$
– Abdullah AKKURT
14 hours ago
1
$begingroup$
@AbdullahAKKURT But with $144$ the answer is an irrational number. If the solution is that straightforward, there is no need to complicate things.
$endgroup$
– Toby Mak
14 hours ago
add a comment |
$begingroup$
A logical solution, however, is given as 144 instead of 441. Otherwise the solution is very easy.
$endgroup$
– Abdullah AKKURT
14 hours ago
1
$begingroup$
@AbdullahAKKURT But with $144$ the answer is an irrational number. If the solution is that straightforward, there is no need to complicate things.
$endgroup$
– Toby Mak
14 hours ago
$begingroup$
A logical solution, however, is given as 144 instead of 441. Otherwise the solution is very easy.
$endgroup$
– Abdullah AKKURT
14 hours ago
$begingroup$
A logical solution, however, is given as 144 instead of 441. Otherwise the solution is very easy.
$endgroup$
– Abdullah AKKURT
14 hours ago
1
1
$begingroup$
@AbdullahAKKURT But with $144$ the answer is an irrational number. If the solution is that straightforward, there is no need to complicate things.
$endgroup$
– Toby Mak
14 hours ago
$begingroup$
@AbdullahAKKURT But with $144$ the answer is an irrational number. If the solution is that straightforward, there is no need to complicate things.
$endgroup$
– Toby Mak
14 hours ago
add a comment |
$begingroup$
My solution
begineqnarray*
3^1 &<&3^x-2<3^2 \
1 &<&x-2<2 \
3 &<&x<4 \
4 &<&2x-2<6 \
frac16 &<½x-2<frac14.
endeqnarray*
Thus,
begineqnarray*
144^frac16 &<&144^frac12x-2<144^frac14 \
2.28943... &<&144^frac12x-2<3.46410...
endeqnarray*
The logical answer is $mathbf3$.
answered 14 hours ago
Abdullah AKKURTAbdullah AKKURT
13
New contributor
Abdullah AKKURT is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
My solution
begineqnarray*
3^1 &<&3^x-2<3^2 \
1 &<&x-2<2 \
3 &<&x<4 \
4 &<&2x-2<6 \
frac16 &<½x-2<frac14.
endeqnarray*
Thus,
begineqnarray*
144^frac16 &<&144^frac12x-2<144^frac14 \
2.28943... &<&144^frac12x-2<3.46410...
endeqnarray*
The logical answer is $mathbf3$.
answered 14 hours ago
Abdullah AKKURTAbdullah AKKURT
13
New contributor
Abdullah AKKURT is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
My solution
begineqnarray*
3^1 &<&3^x-2<3^2 \
1 &<&x-2<2 \
3 &<&x<4 \
4 &<&2x-2<6 \
frac16 &<½x-2<frac14.
endeqnarray*
Thus,
begineqnarray*
144^frac16 &<&144^frac12x-2<144^frac14 \
2.28943... &<&144^frac12x-2<3.46410...
endeqnarray*
The logical answer is $mathbf3$.
answered 14 hours ago
Abdullah AKKURTAbdullah AKKURT
13
New contributor
Abdullah AKKURT is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
My solution
begineqnarray*
3^1 &<&3^x-2<3^2 \
1 &<&x-2<2 \
3 &<&x<4 \
4 &<&2x-2<6 \
frac16 &<½x-2<frac14.
endeqnarray*
Thus,
begineqnarray*
144^frac16 &<&144^frac12x-2<144^frac14 \
2.28943... &<&144^frac12x-2<3.46410...
endeqnarray*
The logical answer is $mathbf3$.
answered 14 hours ago
Abdullah AKKURTAbdullah AKKURT
13
New contributor
Abdullah AKKURT is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 14 hours ago
Abdullah AKKURTAbdullah AKKURT
13
New contributor
Abdullah AKKURT is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 14 hours ago
Abdullah AKKURTAbdullah AKKURT
13
answered 14 hours ago
Abdullah AKKURTAbdullah AKKURT
13
13
New contributor
Abdullah AKKURT is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Abdullah AKKURT is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Abdullah AKKURT is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
Abdullah AKKURT is a new contributor. Be nice, and check out our Code of Conduct.
Abdullah AKKURT is a new contributor. Be nice, and check out our Code of Conduct.
Abdullah AKKURT is a new contributor. Be nice, and check out our Code of Conduct.
Abdullah AKKURT is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
Pretty sure the problem is given or copied wrong.
$endgroup$
– Ian
15 hours ago
$begingroup$
$144$ or $441$??
$endgroup$
– E.H.E
15 hours ago
$begingroup$
None of them. Are you sure about the formula ?
$endgroup$
– Claude Leibovici
15 hours ago
$begingroup$
The question is right. There are no errors.
$endgroup$
– Abdullah AKKURT
15 hours ago
2
$begingroup$
@AbdullahAKKURT It's not right. The result is not one of the options given.
$endgroup$
– Eff
15 hours ago