My solution:

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

Make the substitution $3^x=u$




No solution for $3^x=-1$


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

then $x=1$

  • $\begingroup$ 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

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.


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 $\log_b(b) = 1$.

  • $\begingroup$ 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

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