# Thread: Solving Exponential Equations

1. ## Solving Exponential Equations

I need a quick explanation on how to solve $\displaystyle 2^x = 30$, and how to solve small equations like these when the unknown is an exponent?

Thank you so much.

2. Originally Posted by needhelpplease1
I need a quick explanation on how to solve $\displaystyle 2^x = 30$, and how to solve small equations like these when the unknown is an exponent?

Thank you so much.
use logs ...

$\displaystyle \ln(2^x) = \ln(30)$

$\displaystyle x \cdot \ln(2) = \ln(30)$

$\displaystyle x = \frac{\ln(30)}{\ln(2)}$

3. Originally Posted by skeeter
use logs ...

$\displaystyle \ln(2^x) = \ln(30)$

$\displaystyle x \cdot \ln(2) = \ln(30)$

$\displaystyle x = \frac{\ln(30)}{\ln(2)}$
Sorry, but do you have time to explain one quick thing... did you use a calculator to do that, no right? you just knew to do 30/2 which means that x = 15 right?

4. Originally Posted by needhelpplease1
Sorry, but do you have time to explain one quick thing... did you use a calculator to do that, no right? you just knew to do 30/2 which means that x = 15 right?
no ... $\displaystyle x \ne 15$

$\displaystyle x = \frac{\ln(30)}{\ln(2)} \approx 4.907$

5. Originally Posted by needhelpplease1
I need a quick explanation on how to solve $\displaystyle 2^x = 30$, and how to solve small equations like these when the unknown is an exponent?

Thank you so much.
Consider that $\displaystyle a^b = c \implies b = \log_ac$

Therefore $\displaystyle 2^x = 30 \implies x = \log_230$

This will give you the same answer skeeter's solution provides.