SOLUTION: log(base2)log(base3)log(base4)2^n=2 I don't understand how to solve this.

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: log(base2)log(base3)log(base4)2^n=2 I don't understand how to solve this.      Log On


   



Question 174799: log(base2)log(base3)log(base4)2^n=2
I don't understand how to solve this.

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
log(base2)[log(base3)[log(base4)[2^n] = 2
I assume you are taking log(base2) of [log(base3)[log(base4)[2^n]]
That log is 1
------------------------------
[log(base3)[log(base4)[2^n] = 1
I assume you are taking log(base3) of [log(base4)[2^n]]
That log is zero
log(base4)[2^n] = 0
There is no power of 4 that gives you zero;
So either there is no solution of I am not
interpreting your posting properly.
==============================
Cheers,
Stan H.