<s>[INST] Simplify $\frac{7^{ \log_{5} 15 }+3^{2+\log_{5}7}}{7^{\log_{5}3}}$ I know that the result of this expression is 16 but how do I get to that result?
$$\frac{7^{ \log_{5} 15    }+3^{2+\log_{5}7}}{7^{\log_{5}3}}$$
 [/INST] We simplify the sum in two parts. First notice that $\displaystyle{\frac{7^{\log_5 15}}{7^{\log_5 5}} = 7^{\log_5 15 - \log_5 3} = 7^{\log_5 5} = 7}$. Next rewrite $7^{\log_5 3} = 3^{\log_3 7 \log_5 3}$, so the second part becomes $3^{2 + \log_5 7 - \log_3 7 \log_5 3}$, but $\log_3 7 \log_5 3 = \log_5 3^{\log_3 7} = \log_5 7$ so the second part is equivalent to $3^{2 + \log_5 7 - \log_5 7} = 9$.
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