Question

Solve the following: LOG (8x+1) = 1 + LOG (X-4)

I know the answer is 41/2. Just looking for the steps.

Answer 1

log(8x+1)-log(x-4)=1
log([8x+1]/[x-4])=1
10^ 10^
(8x+1)/(x-4)=10
8x+1=10(x-4)
8x+1=10x-40
1+40=10x-8x
41=2x
x=41/2

Answer 2

Alternate solution:
\[\log (8x+1)=1+\log(x-4) \]
\[\log (8x+1)=\log(10) +\log(x-4) \]
\[\log (8x+1)=\log(10(x-4)) \]
\[8x+1=10x-40\]
\[41=2x\]
\[x={41 \over 2}\]