Take this one for example
1/3log7^x +13log7^y - 5log5^z
How would I reduce this to a single logarithm?
the number the is multiplies with the Logarithm is a power for what is inside it
see this
$\displaystyle a(log(b))=log(a^b)$
$\displaystyle a(log(b)^c)=log(b^{ac})$
so
$\displaystyle \frac{1}{3} log 7^x = log 7^{\frac{x}{3}}$
$\displaystyle 13 log 7^y = log 7^{13y}$
$\displaystyle -5log 5^z=-log 5^{5z}$
now
$\displaystyle log 7^{\frac{x}{3}}+log 7^{13y}-log 5^{5z}=log\left(7^{\frac{x}{3}}\right)\left(7^{13y }\right)-log 5^{5z}$
$\displaystyle log 7^{\frac{x}{3}}+log 7^{13y}-log 5^{5z}=log\dfrac{\left(7^{\frac{x}{3}}\right)\left (7^{13y}\right)}{5^{5z}}$
If those numbers after "log" are the base (ie: $\displaystyle \frac{1}{3}log_7(x) + 13log_7(y) - 5log_5(z)$ then you can combine the first two terms using the laws written by Amer
$\displaystyle log_7(x^{\frac{1}{3}}y^{13})$
Use the change of base rule on this new term and the second term to make them the same base
change bases in logarithms
$\displaystyle log_{a}b=\frac{log_c b}{log_c a}$
so you can write
$\displaystyle -5log_{5}z=-log_5 z^5=-\frac{log_{7}(z^5)}{log_{7}(5)}$
to change it base so you can apply what I mention before
$\displaystyle log_{7}(x^{\frac{1}{3}}y^{13})-\frac{log_{7}(z^5)}{\log_{7}(5)}$
$\displaystyle {\log_{7}(5)}$ consider as constant right
$\displaystyle log_{7}(x^{\frac{1}{3}}y^{13})-log_{7}(z^5)(\log_{7}(5))^{-1}$
$\displaystyle log_{7}(x^{\frac{1}{3}}y^{13})-log_{7}(z^{5(\log_{7}(5))^{-1}})$
$\displaystyle log_{7}\left(\frac{(x^{\frac{1}{3}}y^{13})}{(z^{5( \log_{7}(5))^{-1}})}\right)$