Mixing
Exponents and Log
Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
$3^2x-2left(3^xright)=3$
My solution:
$left(3^xright)^2-2cdot :3^x=3$
Make the substitution $3^x=u$
$left(uright)^2-2u=3$
$u^2-2u-3=0$
$u=3,:u=-1$
No solution for $3^x=-1$
$3^x=3$
I was wondering how i can do this question using logarithms?
then $x=1$
logarithms exponential-function
asked 1 hour ago
HAC
1549
add a comment |Â
up vote
1
down vote
favorite
$3^2x-2left(3^xright)=3$
My solution:
$left(3^xright)^2-2cdot :3^x=3$
Make the substitution $3^x=u$
$left(uright)^2-2u=3$
$u^2-2u-3=0$
$u=3,:u=-1$
No solution for $3^x=-1$
$3^x=3$
I was wondering how i can do this question using logarithms?
then $x=1$
logarithms exponential-function
asked 1 hour ago
HAC
1549
If $3^x=3$, then $x$ is a logarithm. More generally, if $3^x=u$, then $x = log_3(u)$.
– steven gregory
1 hour ago
i want to solve the question using logs from the start. would it work if i take log on both sides?
– HAC
1 hour ago
Are you saying you want to use logarithms from the very beginning? I think your solution is fine as it is, albiet noticing that $3^x = 3 implies log_3(3^x) = log_3(3) implies x = 1$.
– TrostAft
1 hour ago
3
No. Logs haver nice properties over multiplicatin and exponentiation, but not over addition.
– steven gregory
1 hour ago
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
$3^2x-2left(3^xright)=3$
My solution:
$left(3^xright)^2-2cdot :3^x=3$
Make the substitution $3^x=u$
$left(uright)^2-2u=3$
$u^2-2u-3=0$
$u=3,:u=-1$
No solution for $3^x=-1$
$3^x=3$
I was wondering how i can do this question using logarithms?
then $x=1$
logarithms exponential-function
asked 1 hour ago
HAC
1549
$3^2x-2left(3^xright)=3$
My solution:
$left(3^xright)^2-2cdot :3^x=3$
Make the substitution $3^x=u$
$left(uright)^2-2u=3$
$u^2-2u-3=0$
$u=3,:u=-1$
No solution for $3^x=-1$
$3^x=3$
I was wondering how i can do this question using logarithms?
then $x=1$
logarithms exponential-function
logarithms exponential-function
asked 1 hour ago
HAC
1549
asked 1 hour ago
HAC
1549
asked 1 hour ago
HAC
1549
asked 1 hour ago
HAC
1549
asked 1 hour ago
HAC
1549
1549
If $3^x=3$, then $x$ is a logarithm. More generally, if $3^x=u$, then $x = log_3(u)$.
– steven gregory
1 hour ago
i want to solve the question using logs from the start. would it work if i take log on both sides?
– HAC
1 hour ago
Are you saying you want to use logarithms from the very beginning? I think your solution is fine as it is, albiet noticing that $3^x = 3 implies log_3(3^x) = log_3(3) implies x = 1$.
– TrostAft
1 hour ago
3
No. Logs haver nice properties over multiplicatin and exponentiation, but not over addition.
– steven gregory
1 hour ago
add a comment |Â
If $3^x=3$, then $x$ is a logarithm. More generally, if $3^x=u$, then $x = log_3(u)$.
– steven gregory
1 hour ago
i want to solve the question using logs from the start. would it work if i take log on both sides?
– HAC
1 hour ago
Are you saying you want to use logarithms from the very beginning? I think your solution is fine as it is, albiet noticing that $3^x = 3 implies log_3(3^x) = log_3(3) implies x = 1$.
– TrostAft
1 hour ago
3
No. Logs haver nice properties over multiplicatin and exponentiation, but not over addition.
– steven gregory
1 hour ago
If $3^x=3$, then $x$ is a logarithm. More generally, if $3^x=u$, then $x = log_3(u)$.
– steven gregory
1 hour ago
If $3^x=3$, then $x$ is a logarithm. More generally, if $3^x=u$, then $x = log_3(u)$.
– steven gregory
1 hour ago
i want to solve the question using logs from the start. would it work if i take log on both sides?
– HAC
1 hour ago
i want to solve the question using logs from the start. would it work if i take log on both sides?
– HAC
1 hour ago
Are you saying you want to use logarithms from the very beginning? I think your solution is fine as it is, albiet noticing that $3^x = 3 implies log_3(3^x) = log_3(3) implies x = 1$.
– TrostAft
1 hour ago
Are you saying you want to use logarithms from the very beginning? I think your solution is fine as it is, albiet noticing that $3^x = 3 implies log_3(3^x) = log_3(3) implies x = 1$.
– TrostAft
1 hour ago
3
3
No. Logs haver nice properties over multiplicatin and exponentiation, but not over addition.
– steven gregory
1 hour ago
No. Logs haver nice properties over multiplicatin and exponentiation, but not over addition.
– steven gregory
1 hour ago
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
To answer your specific inquiry:
I was wondering how i can do this question using logarithms?
I guess your asking if a solution would exist by taking the log of both sides. If this is what you mean, then I would say it is is possible but is not simple. Your solution is better. The expression:
$log(x-y) = log (z)$
Is not an expression you can manipulate further with ease.
Remember that:
$(log(x-y))$ is NOT always equal to ($log (x) - log (y))$
In fact, the equality only holds for specific values of y and x:
As a result, since you can't simplify the original problem by taking the log of both sides, your answer is good.
answered 1 hour ago
NoChance
3,38321221
add a comment |Â
up vote
2
down vote
To solve $3^x = 3$ using logs, you can just take the logarithm of both sides (with respect to any base). Typically you can get away with always taking logs of both sides with base $e$ (these are called natural logs and are denoted $log_e(x) = ln(x)$), or in this case you apply $log_3 (cdot )$ to both sides:
$$
3^x = 3 implies log_3(3^x) = log_3(3) implies x = 3
$$
where we used that $log_b(x^p) = p log_b(x)$ and $log_b(b) = 1$.
answered 1 hour ago
JavaMan
10.7k12653
i know how to solve $3^x = 3$. i want to solve it using logs from the start
– HAC
1 hour ago
1
Your method was the "right" way to solve it from the start. You mainly use logarithms to solve exponential equations like $a^x = b$.
– JavaMan
1 hour ago
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
To answer your specific inquiry:
I was wondering how i can do this question using logarithms?
I guess your asking if a solution would exist by taking the log of both sides. If this is what you mean, then I would say it is is possible but is not simple. Your solution is better. The expression:
$log(x-y) = log (z)$
Is not an expression you can manipulate further with ease.
Remember that:
$(log(x-y))$ is NOT always equal to ($log (x) - log (y))$
In fact, the equality only holds for specific values of y and x:
As a result, since you can't simplify the original problem by taking the log of both sides, your answer is good.
answered 1 hour ago
NoChance
3,38321221
add a comment |Â
up vote
2
down vote
accepted
To answer your specific inquiry:
I was wondering how i can do this question using logarithms?
I guess your asking if a solution would exist by taking the log of both sides. If this is what you mean, then I would say it is is possible but is not simple. Your solution is better. The expression:
$log(x-y) = log (z)$
Is not an expression you can manipulate further with ease.
Remember that:
$(log(x-y))$ is NOT always equal to ($log (x) - log (y))$
In fact, the equality only holds for specific values of y and x:
As a result, since you can't simplify the original problem by taking the log of both sides, your answer is good.
answered 1 hour ago
NoChance
3,38321221
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
To answer your specific inquiry:
I was wondering how i can do this question using logarithms?
I guess your asking if a solution would exist by taking the log of both sides. If this is what you mean, then I would say it is is possible but is not simple. Your solution is better. The expression:
$log(x-y) = log (z)$
Is not an expression you can manipulate further with ease.
Remember that:
$(log(x-y))$ is NOT always equal to ($log (x) - log (y))$
In fact, the equality only holds for specific values of y and x:
As a result, since you can't simplify the original problem by taking the log of both sides, your answer is good.
answered 1 hour ago
NoChance
3,38321221
To answer your specific inquiry:
I was wondering how i can do this question using logarithms?
I guess your asking if a solution would exist by taking the log of both sides. If this is what you mean, then I would say it is is possible but is not simple. Your solution is better. The expression:
$log(x-y) = log (z)$
Is not an expression you can manipulate further with ease.
Remember that:
$(log(x-y))$ is NOT always equal to ($log (x) - log (y))$
In fact, the equality only holds for specific values of y and x:
As a result, since you can't simplify the original problem by taking the log of both sides, your answer is good.
answered 1 hour ago
NoChance
3,38321221
answered 1 hour ago
NoChance
3,38321221
answered 1 hour ago
NoChance
3,38321221
answered 1 hour ago
NoChance
3,38321221
3,38321221
add a comment |Â
add a comment |Â
up vote
2
down vote
To solve $3^x = 3$ using logs, you can just take the logarithm of both sides (with respect to any base). Typically you can get away with always taking logs of both sides with base $e$ (these are called natural logs and are denoted $log_e(x) = ln(x)$), or in this case you apply $log_3 (cdot )$ to both sides:
$$
3^x = 3 implies log_3(3^x) = log_3(3) implies x = 3
$$
where we used that $log_b(x^p) = p log_b(x)$ and $log_b(b) = 1$.
answered 1 hour ago
JavaMan
10.7k12653
i know how to solve $3^x = 3$. i want to solve it using logs from the start
– HAC
1 hour ago
1
Your method was the "right" way to solve it from the start. You mainly use logarithms to solve exponential equations like $a^x = b$.
– JavaMan
1 hour ago
add a comment |Â
up vote
2
down vote
To solve $3^x = 3$ using logs, you can just take the logarithm of both sides (with respect to any base). Typically you can get away with always taking logs of both sides with base $e$ (these are called natural logs and are denoted $log_e(x) = ln(x)$), or in this case you apply $log_3 (cdot )$ to both sides:
$$
3^x = 3 implies log_3(3^x) = log_3(3) implies x = 3
$$
where we used that $log_b(x^p) = p log_b(x)$ and $log_b(b) = 1$.
answered 1 hour ago
JavaMan
10.7k12653
i know how to solve $3^x = 3$. i want to solve it using logs from the start
– HAC
1 hour ago
1
Your method was the "right" way to solve it from the start. You mainly use logarithms to solve exponential equations like $a^x = b$.
– JavaMan
1 hour ago
add a comment |Â
up vote
2
down vote
up vote
2
down vote
To solve $3^x = 3$ using logs, you can just take the logarithm of both sides (with respect to any base). Typically you can get away with always taking logs of both sides with base $e$ (these are called natural logs and are denoted $log_e(x) = ln(x)$), or in this case you apply $log_3 (cdot )$ to both sides:
$$
3^x = 3 implies log_3(3^x) = log_3(3) implies x = 3
$$
where we used that $log_b(x^p) = p log_b(x)$ and $log_b(b) = 1$.
answered 1 hour ago
JavaMan
10.7k12653
To solve $3^x = 3$ using logs, you can just take the logarithm of both sides (with respect to any base). Typically you can get away with always taking logs of both sides with base $e$ (these are called natural logs and are denoted $log_e(x) = ln(x)$), or in this case you apply $log_3 (cdot )$ to both sides:
$$
3^x = 3 implies log_3(3^x) = log_3(3) implies x = 3
$$
where we used that $log_b(x^p) = p log_b(x)$ and $log_b(b) = 1$.
answered 1 hour ago
JavaMan
10.7k12653
answered 1 hour ago
JavaMan
10.7k12653
answered 1 hour ago
JavaMan
10.7k12653
answered 1 hour ago
JavaMan
10.7k12653
10.7k12653
i know how to solve $3^x = 3$. i want to solve it using logs from the start
– HAC
1 hour ago
1
Your method was the "right" way to solve it from the start. You mainly use logarithms to solve exponential equations like $a^x = b$.
– JavaMan
1 hour ago
add a comment |Â
i know how to solve $3^x = 3$. i want to solve it using logs from the start
– HAC
1 hour ago
1
Your method was the "right" way to solve it from the start. You mainly use logarithms to solve exponential equations like $a^x = b$.
– JavaMan
1 hour ago
i know how to solve $3^x = 3$. i want to solve it using logs from the start
– HAC
1 hour ago
i know how to solve $3^x = 3$. i want to solve it using logs from the start
– HAC
1 hour ago
1
1
Your method was the "right" way to solve it from the start. You mainly use logarithms to solve exponential equations like $a^x = b$.
– JavaMan
1 hour ago
Your method was the "right" way to solve it from the start. You mainly use logarithms to solve exponential equations like $a^x = b$.
– JavaMan
1 hour ago
add a comment |Â
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If $3^x=3$, then $x$ is a logarithm. More generally, if $3^x=u$, then $x = log_3(u)$.
– steven gregory
1 hour ago
i want to solve the question using logs from the start. would it work if i take log on both sides?
– HAC
1 hour ago
Are you saying you want to use logarithms from the very beginning? I think your solution is fine as it is, albiet noticing that $3^x = 3 implies log_3(3^x) = log_3(3) implies x = 1$.
– TrostAft
1 hour ago
3
No. Logs haver nice properties over multiplicatin and exponentiation, but not over addition.
– steven gregory
1 hour ago