Exponents and Log

Question

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

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 $x=1$

Answer 1

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.

Answer 2

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 $\log_b(b) = 1$.