How do I solve logarithmic expressions that have different bases My math teacher asked me to simplify the following expression to a single logarithm and then evaluate. 
$$\log_{5}125 + \log_{25}5 - 3\log(4)$$  
The bases are different and I found it quite hard to express them as a single log. On my working out I changed the base of $$\log_{25}5$$ to $$\log_{5}25$$ 
$$\log_{5}125 + \log_{5}25 - \log4^3$$ $$\log_{5}(125×25)-\log64$$ $$\log_{5}(3125)-\log64$$  That's all I could simplify it to. 
 A: You have incorrectly changed the base.
A useful logarithm rule is the 'change of base rule'. It states:
$$\log_ab=\frac{\log_na}{\log_nb}$$
This lets you change any base to any other base as the $n$ only appears on the right hand side.
In questions like the one you teacher has given simplification can occur when you have a base which is a rational power of another base. In your problem we see that $25$ is a power of $5$ so we can simplify the second terms as follows:
$$\log_{25}{5}=\frac{\log_55}{\log_525}$$
$$=\frac{1}{\log_55^2}$$
$$=\frac{1}{2\log_55}$$
$$=\frac{1}{2}$$
You can then add it to the first term which you should be able to simplify to 3 by basic definitions.
So this would give you an answer of $$3+\frac{1}{2}-\log_{10}64$$
Additional:
Your textbook asked for it as a single logarithm. This is doable but won't look pretty so I'm questioning if that is actually what the book meant.
$$3+\frac{1}{2}-\log_{10}64=\frac{7}{2}-\log_{10}64$$
$$=\frac{7}{2}\log_{10}{10}-\frac{2}{2}\log_{10}64$$
$$=\frac{1}{2}\left(7\log_{10}10-2\log_{10}64\right)$$
$$=\frac{1}{2}\left(\log_{10}10^7-\log_{10}64^2\right)$$
$$=\frac{1}{2}\left(\log_{10}\frac{10^7}{4096}\right)$$
$$=\frac{1}{2}\left(\log_{10}\frac{78125}{32}\right)$$
$$=\log_{10}\sqrt{\frac{78125}{32}}$$
