Exponents and Log $3^{2x}-2\left(3^x\right)=3$
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 $x=1$ 
 A: 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.
A: 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 $\log_b(b) = 1$.
