SOLUTION: How do you solve a log equations where there is x on both sides and one where there is only one x on one side? Such as the following? {{{ log ( 3, x )^2}}} ={{{ log( 3, 4 )}}}

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: How do you solve a log equations where there is x on both sides and one where there is only one x on one side? Such as the following? {{{ log ( 3, x )^2}}} ={{{ log( 3, 4 )}}}       Log On


   



Question 30396: How do you solve a log equations where there is x on both sides and one where there is only one x on one side? Such as the following?
+log+%28+3%2C+x+%29%5E2 =+log%28+3%2C+4+%29
&
+log+%282%2C+%28x%2B2%29%29%5E2 = log+%282%2C%283x%2B16%29%29
Thank you very much!!!

Found 2 solutions by Paul, josmiceli:
Answer by Paul(988) About Me  (Show Source):
You can put this solution on YOUR website!
a) Since logarithm's bases are equal they can cancel out:




b)
x%5E2%2B4x%2B4=3x%2B16
x%5E2%2Bx-12=0
Factor: (x+4) or (x-3)
x=-4 or x=3
Remove negative x =3
Paul.

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
+log+%28+3%2C+x+%29%5E2 =+log%28+3%2C+4+%29
I always say "logs are exponents" before I start with these
Then I think about what the equation says
The left side says there is an exponent (the right side) to
the base 3 that gives me x^2
3^(right side) = x^2
3%5E%28log+%283%2C4%29%29+=+x%5E2
You have to say the left side to get it
The left side says "3 raised to the exponent to the base 3 that
gives me 4"
All that is just 4 if it gives me 4, so
4+=+x%5E2
x+=+%2B2
x+=+-2
----------------------
+log+%282%2C+%28x%2B2%29%29%5E2 = log+%282%2C%283%2Ax%2B16%29%29
same thing
2^(right side) = (x+2)^2
2%5Elog%282%2C%283%2Ax%2B16%29%29+=+%28x%2B2%29%5E2
same thing again
It says "raise 2 to an exponent to the base 2 that gives me 3x + 16"
That's just 3x + 16
3%2Ax+%2B+16+=+%28x+%2B+2%29%5E2
3%2Ax+%2B+16+=+X%5E2+%2B+4%2Ax+%2B+4
x%5E2+%2B+x+-+12+=+0
%28x+-+3%29%28x+%2B+4%29+=+0
x+=+3
y+=+-4