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)

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)      Log On


   

Question 829852: 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)
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
find x
1) log%282%2C%28x-1%29%29+%2B+log%282%2C%28x%2B1%29%29 = 3
we can write this as single log
log%282%2C%28%28x-1%29%28x%2B1%29%29%29 = 3
log%282%2C%28x%5E2-1%29%29 = 3
write the exponent equiv of logs
%28x%5E2+-+1%29+=+2%5E3
%28x%5E2+-+1%29+=+8
x%5E2+-+1+-+8+=+0
x%5E2+-+9 = 0
Factors as the difference of squares
(x+3)(x-3) = 0
x = +3, the only valid solution, can't have the log of a neg value
:
:
2) log (4-2x)=log(x+1)
if the logs are equal the expressions are equal, just write it:
4 - 2x = x + 1
4 - 1 = x + 2x
3 = 3x
x = 1
:
:
3) log%281%2F3%2C%289-2x%29%29 =log%281%2F3%2C%28x-6%29%29
If the expressions are equal of logs with same base
9 - 2x = x - 6
9 + 6 = x + 2x
15 = 3x
x = 5
However if you substitute in the original problem, we end up with -1 on each
side, no log of neg number is allowed, so no solution