# log base 2 help

• Jul 11th 2007, 02:00 PM
davecs77
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.
• Jul 11th 2007, 02:09 PM
red_dog
We have \$\displaystyle \log_aa^n=n\$.
So \$\displaystyle \log_22^{4x}=20\Leftrightarrow 4x=20\Leftrightarrow x=5\$
• Jul 11th 2007, 02:12 PM
janvdl
Quote:

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. :confused:
• Jul 11th 2007, 02:17 PM
davecs77
Quote:

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?
• Jul 11th 2007, 02:19 PM
janvdl
Quote:

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 \$