1
$\begingroup$

$3^{2x}-2\left(3^x\right)=3$

My solution:

$\left(3^x\right)^2-2\cdot \:3^x=3$

Make the substitution $3^x=u$

$\left(u\right)^2-2u=3$

$u^2-2u-3=0$

$u=3,\:u=-1$

No solution for $3^x=-1$

$3^x=3$

I was wondering how i can do this question using logarithms?

then $x=1$

$\endgroup$
4
  • $\begingroup$ If $3^x=3$, then $x$ is a logarithm. More generally, if $3^x=u$, then $x = \log_3(u)$. $\endgroup$ Oct 15 '18 at 2:50
  • $\begingroup$ i want to solve the question using logs from the start. would it work if i take log on both sides? $\endgroup$
    – HAC
    Oct 15 '18 at 2:52
  • $\begingroup$ Are you saying you want to use logarithms from the very beginning? I think your solution is fine as it is, albiet noticing that $3^x = 3 \implies \log_3(3^x) = \log_3(3) \implies x = 1$. $\endgroup$
    – TrostAft
    Oct 15 '18 at 2:56
  • 3
    $\begingroup$ No. Logs haver nice properties over multiplicatin and exponentiation, but not over addition. $\endgroup$ Oct 15 '18 at 2:57
4
$\begingroup$

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:

enter image description here

As a result, since you can't simplify the original problem by taking the log of both sides, your answer is good.

$\endgroup$
3
$\begingroup$

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 $\log_b(b) = 1$.

$\endgroup$
2
  • $\begingroup$ i know how to solve $3^x = 3$. i want to solve it using logs from the start $\endgroup$
    – HAC
    Oct 15 '18 at 2:56
  • 2
    $\begingroup$ Your method was the "right" way to solve it from the start. You mainly use logarithms to solve exponential equations like $a^x = b$. $\endgroup$
    – JavaMan
    Oct 15 '18 at 2:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.