SOLUTION: could you please show me how to slove these to find x. log(x+1)+ log(x-1)=log24 2log(base of 4)x-log(base of 4)(x+3)=1

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: could you please show me how to slove these to find x. log(x+1)+ log(x-1)=log24 2log(base of 4)x-log(base of 4)(x+3)=1      Log On


   

Question 679707: could you please show me how to slove these to find x.
log(x+1)+ log(x-1)=log24
2log(base of 4)x-log(base of 4)(x+3)=1
Answer by pmatei(79) About Me  (Show Source):
You can put this solution on YOUR website!
log%28x%2B1%29+%2B+log%28x-1%29+=+log24
log%28%28x%2B1%29%28x-1%29%29+=+log24
%28x%2B1%29%28x-1%29+=+24
x%5E2+-+1+=+24
x%5E2+=+25
x+=+5 or x=-5
You cannot have negative numbers under logarithm, so the only solution is:
x+=+5

2log%284%2Cx%29+-+log%284%2C%28x%2B3%29%29+=+1
log%284%2Cx%5E2%29+-+log%284%2C%28x%2B3%29%29+=+log%284%2C4%29
log%284%2C%28x%5E2%2F%28x%2B3%29%29%29+=+log%284%2C4%29
x%5E2%2F%28x%2B3%29+=+4
Because of the (x+3) in the denominator of the fractions, x cannot be -3 (we cannot divide by zero).
x%5E2+=+4%28x%2B3%29
x%5E2+=+4x+%2B12
x%5E2+-+4x+-+12+=+0
%28x-6%29%28x%2B2%29+=+0
x-6+=+0 or x%2B2=0
x=6 or x=-2
Because of the log%284%2Cx%29, we cannot accept the x=-2 solution - no negative numbers under logarithm.
Final solution x=6.