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 $$x=1$$

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

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.

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 $$\log_b(b) = 1$$.

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