logarithmic equations log2x-log(x-7)=log3

Question

log2x-log(x-7)=log3

just not sure how to solve it when theres a log on each side with a different log number :)

Answer 1

We know that logab = loga+logb

loga/b = loga-logb

Given equation log2x -log(x-7) = log3

Subtract log3 from each side.

log2x -log(x-7)- log3 = log3-log3

log2x-(log(x-7)+log3) = 0

log2x-(log3(x-7)) = 0

log2x/3(x-7) = 0

log2x/3(x-7) = log1

2x/3(x-7) = 1

Multple to each side by 3(x-7).

2x/3(x-7)*3(x-7). = 1*3(x-7).

2x = 3(x-7)

2x = 3x-21

Subtract 3x from each side.

2x-3x = -21+3x-3x

-x = -21

Multple to each side by negitive one.

Solution x = 21