IF you know the "laws of logarithms": a log(b)= log(b^a) and log(a)+ log(b)= log(ab), this problem is easy. If you do NOT the you should not be attempting a problem like this.
You are assuming that an unqualified $\displaystyle \log$ denotes $\displaystyle \log_{10}$ which might be the case in high school maths classes but is not in the wider world.
What are the real solutions, in any, of log(x) = 2log(3) + log(5) - log(2) - 1 (all logs are to base 10)
Originally Posted by zzephod
You are assuming that an unqualified $\displaystyle \log$ denotes $\displaystyle \log_{10}$ which might be the case in high school maths classes but is not in the wider world..
The OP clearly states they are indeed base 10.
Please read the entire thread before positing.