# Thread: log base 2 help

1. ## log base 2 help

I have to find x in this problem

logbase2(2^4x) = 20

The 2 is raised to the power 4x.

I started off by doing (2^4x)/ln2 = 20
then got 2^4x = 20ln2 and got stuck too.
Thanks.. I am trying to get ready for my quiz tommorow in calc 2.
Thanks for all the help.

2. We have $\displaystyle \log_aa^n=n$.
So $\displaystyle \log_22^{4x}=20\Leftrightarrow 4x=20\Leftrightarrow x=5$

3. Originally Posted by red_dog
We have $\displaystyle \log_aa^n=n$.
So $\displaystyle \log_22^{4x}=20\Leftrightarrow 4x=20\Leftrightarrow x=5$
I solved it using the same method, but i thought i may have missed something because it's supposed to be university maths.

4. Originally Posted by red_dog
We have $\displaystyle \log_aa^n=n$.
So $\displaystyle \log_22^{4x}=20\Leftrightarrow 4x=20\Leftrightarrow x=5$

How did you get from logbase2 2^(4x) = 20 to 4x = 20?

5. Originally Posted by davecs77
How did you get from logbase2 2^(4x) = 20 to 4x = 20?
$\displaystyle log_{2} 2^{4x} = 20$

$\displaystyle 4x \ log_{2} 2 = 20$

$\displaystyle But \ log_{2} 2 = 1$

$\displaystyle 4x (1) = 20$

$\displaystyle x = 5$