SOLUTION: Solve for x: logbase2(logbase3(logbase4 x)=0

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: Solve for x: logbase2(logbase3(logbase4 x)=0      Log On


   

Question 134967: Solve for x: logbase2(logbase3(logbase4 x)=0
Found 2 solutions by stanbon, josmiceli:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
Solve for x: logbase2(logbase3(logbase4 x)=0
----------------------------
work from the far left to the right:
= log3(log4 x) = 2^0 = 1
=log4 x = 3^1 = 3
x = 4^3 = 64
=================
Cheers,
Stan H.

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
log%282%2C%28log%283%2C%28log%284%2Cx%29%29%29%29%29+=+0
Just looking at the outside pair of parenthses and nothing
inside, I see log%282%2C+k%29+=+0 where k is all the other
stuff. This is telling me 2%5E0+=+k or k+=+1, so
log%283%2C%28log%284%2Cx%29%29%29+=+1
Now I can say log%283%2Cj%29+=+1, where j+=+log%284%2Cx%29
This tells me 3%5E1+=+j or j+=+3
Now I have
log%284%2Cx%29+=+3 or 4%5E3+=+x
x+=+64 answer