log(3m-5) + log(m) = log(2)

Use rule: log(A) + log(B) = log(AB) on the
left side

log[(3m-5)·m] = log(2)

  log[3m²-5m] = log(2)

Use rule: log(A) = log(B) is equivalent to A = B

      3m²-5m = 2

can you solve that quadratic equation. If not
post again

  m = -1/3, 2

The log of a negative number is not defined. When
we substitute -1/3 for m in

log(3m-5) + log(m) = log(2)

the left side is undefined.
So we discard -1/3.

Checking m = 2

log(3·2-5) + log(2) = log(2)
  log(6-5) + log(2) = log(2)
    log(1) + log(2) = log(2)
         0 + log(2) = log(2)
             log(2) = log(2)

So m = 2 is the only solution.  

Edwin