SOLUTION: Solve the following equations for x: {{{ 3^( log ( 5, x))^2 = 3^( 12 + log (5, x)) }}}

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: Solve the following equations for x: {{{ 3^( log ( 5, x))^2 = 3^( 12 + log (5, x)) }}}      Log On


   

Question 872836: Solve the following equations for x:
+3%5E%28+log+%28+5%2C+x%29%29%5E2+=+3%5E%28+12+%2B+log+%285%2C+x%29%29+
Found 2 solutions by psbhowmick, MathTherapy:
Answer by psbhowmick(878) About Me  (Show Source):
You can put this solution on YOUR website!
+3%5E%28+log+%28+5%2C+x%29%29%5E2+=+3%5E%28+12+%2B+log+%285%2C+x%29%29+
Let u+=+log+%285%2C+x%29.

+3%5E%28u%5E2%29+=+3%5E%28+12+%2B+u%29+
+3%5E%28u%5E2%29+%2F+3%5E%28+12+%2B+u%29+=+1
+3%5E%28u%5E2+-+12+-+u%29+=+1
u%5E2+-+12+-+u+=+0
%28u+-+4%29%28u+%2B+3%29+=+0
u+=+4 or u+=+-3
log+%285%2C+x%29+=+4 or log+%285%2C+x%29+=+-3
x+=+5%5E4 or x+=+5%5E%28-3%29
x+=+625 or x+=+1%2F125

Answer by MathTherapy(10555) About Me  (Show Source):
You can put this solution on YOUR website!
Solve the following equations for x:
+3%5E%28+log+%28+5%2C+x%29%29%5E2+=+3%5E%28+12+%2B+log+%285%2C+x%29%29+


Assuming the problem is: +3%5E%28+log+%28+5%2C+x%5E2%29%29+=+3%5E%28+12+%2B+log+%285%2C+x%29%29+, with the bases being the same, their exponents are equal

Thus, log%285%2C+x%5E2%29+=+12+%2B+log%285%2C+x%29

2log%285%2C+x%29+=+12+%2B+log%285%2C+x%29

2log%285%2C+x%29+-+log%285%2C+x%29+=+12

log%285%2C+x%29+=+12 

highlight_green%28x+=+5%5E12%29

You can do the check!! 

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