SOLUTION: solve: {{{log base x[log base 2(log base 3 of 9)]=2}}} Unfortunately I do not know how to put this without all the words. But this is a problem on one of my take home tests. I do

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: solve: {{{log base x[log base 2(log base 3 of 9)]=2}}} Unfortunately I do not know how to put this without all the words. But this is a problem on one of my take home tests. I do      Log On


   

Question 250018: solve:
log+base+x%5Blog+base+2%28log+base+3+of+9%29%5D=2
Unfortunately I do not know how to put this without all the words. But this is a problem on one of my take home tests. I dont understand how to solve because I get 3%5E2=9 for the log in the parentheses but now I don't knwo what to do?
Found 2 solutions by stanbon, MathTherapy:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
log+base+x%5Blog+base+2%28log+base+3+of+9%29%5D=2
----
log3(9) = 2 so your problem becomes:
log+base+x%5Blog+base+2%282%29%5D=2
---
log2(2) = 1 so your problem becomes:
logx(1) = 2
---
x^2 = 1
x^2-1 = 0
(x-1)(x+1) = 0
x = 1 or x=-1
But -1 cannot be the base of a log.
So x = 1 is the only solution.
=================================
Cheers,
Stan H.

Answer by MathTherapy(10699) About Me  (Show Source):
You can put this solution on YOUR website!
solve:
log+base+x%5Blog+base+2%28log+base+3+of+9%29%5D=2
Unfortunately I do not know how to put this without all the words. But this is a problem on one of my take home tests. I dont understand how to solve because I get 3%5E2=9 for the log in the parentheses but now I don't knwo what to do?
===============================================
In this problem, x is one of the 3 bases given. So, being a LOG-BASE, x CANNOT be 0, and also, x MUST be GREATER than 1. 

 log+base+x%5Blog+base+2%28log+base+3+of+9%29%5D=2 
  log+%28x%2C+%28log+%282%2C+%28log+%283%2C+%289%29%29%29%29%29%29+=+2, with x > 1

 log+%28x%2C+%28log+%282%2C+highlight%28%28log+%283%2C+%289%29%29%29%29%29%29%29+=+2
         log+%28x%2C+%28log+%282%2C+%282%29%29%29%29+=+2 ---  log+%283%2C+%289%29%29+=+log+%283%2C+%283%5E2%29%29+=+2
        log+%28x%2C+highlight%28%28log+%282%2C+%282%29%29%29%29%29+=+2
                log+%28x%2C+%281%29%29+=+2 ---  log+%282%2C+%282%29%29+=+1
                    x%5E2+=+1 ----- Converting to EXPONENTIAL form
               x+=+0%2B-+sqrt%281%29+=+0%2B-+1

However, as stated above, x MUST be > 1, so There are NO SOLUTIONS to this equation.

The other person is WRONG, as he/she states that the value of x is 1.