SOLUTION: find x 1) log (x-1) to base 2 + log (x+1) to base 2 = 3 2) log (4-2x)=log(x+1) 3) log (9-2x) to base 1/3=log1/3(x-6) thank you so much!

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: find x 1) log (x-1) to base 2 + log (x+1) to base 2 = 3 2) log (4-2x)=log(x+1) 3) log (9-2x) to base 1/3=log1/3(x-6) thank you so much!      Log On


   

Question 829853: find x
1) log (x-1) to base 2 + log (x+1) to base 2 = 3
2) log (4-2x)=log(x+1)
3) log (9-2x) to base 1/3=log1/3(x-6)
thank you so much!
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
find x
1) log (x-1) to base 2 + log (x+1) to base 2 = 3
log%282%2C%28x-1%29%29+%2B+log%282%2Cx%2B1%29%29+=+3+=+log%282%2C8%29
log%282%2C%28x-1%29%2A%28x%2B1%29%29+=+log%282%2C8%29
%28x-1%29%2A%28x%2B1%29+=+8
x%5E2+-+1+=+8
x%5E2+=+9
x = 3
----
x = -3 is rejected, gives log(-4)
===================================
2) log (4-2x)=log(x+1)
Simple
=============
3) log (9-2x) to base 1/3=log1/3(x-6)
Is 1/3 the base of both?
Use log(1/3,(9-2x)) with 3 sets of braces { } to render legibly.
Click "Show source" to see the format.