<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] Use properties of logs first: 
\begin{align*}
\dfrac{7^{\log_5 15} + 3^{2+\log_5 7}}{7^{\log_5 3}}
= \dfrac{7^{\log_5 5 + \log_5 3} + 3^{2}3^{\log_5 7}}{7^{\log_5 3}}
&= \dfrac{7^{1}7^{\log_5 3} + 3^{2}3^{\log_5 7}}{7^{\log_5 3}} \\
&= 7 \biggl( \dfrac{7^{\log_5 3}}{7^{\log_5 3}}\biggr) + 9 \biggl(\dfrac{3^{\log_5 7}}{7^{\log_5 3}}\biggr) \\
&= 7 + 9\biggl(\dfrac{3^{\log_5 7}}{7^{\log_5 3}}\biggr).
\end{align*}
Now we need to determine the quantity in the parentheses. Put $x=3^{\log_5 7}$ and $y = 7^{\log_5 3}$; then the quantity in the parentheses is $x/y$. We have $$\log_5 x = \log_5 (3^{\log_5 7}) = \log_5 7 \cdot \log_5 3 = \log_5 3 \cdot \log_5 7 = \log_5 (7^{\log_5 3}) = \log_5 y,$$ so $x=y$. 
Therefore, $$\dfrac{7^{\log_5 15} + 3^{2+\log_5 7}}{7^{\log_5 3}} = 7 + 9\biggl(\dfrac{3^{\log_5 7}}{7^{\log_5 3}}\biggr) = 7 + 9\biggl(\dfrac{x}{y}\biggr) = 7+9(1) = 16.$$
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