SOLUTION: {{{2^(logx)=4*x}}} I started with: {{{2^(logx)=2^2*x}}} then got: Logx=2*x don't know where to go from there

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: {{{2^(logx)=4*x}}} I started with: {{{2^(logx)=2^2*x}}} then got: Logx=2*x don't know where to go from there      Log On


   

Question 244851: 2%5E%28logx%29=4%2Ax
I started with:
2%5E%28logx%29=2%5E2%2Ax
then got:
Logx=2*x
don't know where to go from there
Answer by rapaljer(4671) About Me  (Show Source):
You can put this solution on YOUR website!
2%5E%28logx%29=4%2Ax

Take the log of each side:
log%282%5E%28logx%29%29=log%284x%29
logx%2Alog2=log4%2Blogx
Subtract logx from each side:
logx%2Alog2-logx=log4

Factor out the logx:
%28logx%29%28log2+-1%29=log4

Divide both sides by (log2-1):
%28%28logx%29%28log2-1%29%29%2F%28log2-1%29=%28log4%29%2F%28log2-1%29
logx=+%28log4%29%2F%28log2-1%29

Finally, raise both sides as a power of 10:
x=10%5E%28%28log4%29%2F%28log2-1%29%29.

This comes out to approximately .137609.

I checked this with a graphing calculator, and it is correct!!!

Dr. Robert J. Rapalje, Retired
Seminole State College of Florida