SOLUTION: Log(√x+7) + log2 = log(x-1)

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: Log(√x+7) + log2 = log(x-1)      Log On


   

Question 990797: Log(√x+7) + log2 = log(x-1)
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
log%28sqrt%28x%29%2B7%29+%2B+log%282%29+=+log%28x-1%29
log%282%28sqrt%28x%29%2B7%29%29+=+log%28x-1%29....if log same, we have
2%28sqrt%28x%29%2B7%29+=+x-1.......solve for x
2sqrt%28x%29%2B14+=+x-1
2sqrt%28x%29+=+x-1-14
2sqrt%28x%29+=+x-15
%282sqrt%28x%29%29%5E2+=+%28x-15%29%5E2
4x+=+%28x-15%29%5E2
4x+=+x%5E2-30x%2B15%5E2
0=+x%5E2-30x-4x%2B225
0=+x%5E2-34x%2B225
0=+x%5E2-9x-25x%2B225
0=%28+x%5E2-9x%29-%2825x-225%29
0=x%28+x-9%29-25%28x-9%29
0+=+%28x-25%29%28x-9%29
solutions:
if 0+=+%28x-25%29=>x=25
if 0+=+%28x-9%29=>x=9
solution we can use to make a statement log%28sqrt%28x%29%2B7%29+%2B+log%282%29+=+log%28x-1%29 true is =>x=25