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?
$$\frac{7^{ \log_{5} 15    }+3^{2+\log_{5}7}}{7^{\log_{5}3}}$$
 A: Hint:
$$\log_5{15}=1+\log_53$$
If $\log_53=x,3=5^x\implies5^{\log_53}=5^x=3$
$$3^{\log_57}=(5^{\log_53})^{\log_57}=(5^{\log_57})^{\log_53}=7^{\log_53}$$
A: Using $$\boxed{a^{\log_b{c}}=c^{\log_b{a}}}$$
$$7^{\log_5{15}}=15^{\log_5{7}}$$
Simplifying ,
$$\frac{15^{\log_5{7}} + 3^\left({2+\log_5{7}}\right)}{3^{\log_5{7}}}$$
Taking ${\log_5{7}}$ as a variable $t$, we get 
$$\frac{15^t+ 3^\left({2+t}\right)}{3^t}$$
$$\frac{3^t\cdot5^t + 9\cdot3^t}{3^t}$$
which simplifies to 
$$5^t+9$$
Now ,back substituting the value of $t$,
$$5^{\log_5{7}}+9$$
Using the same property of log again ,
$$7^{\log_5{5}}+9=7^1+9=16$$
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A: 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 $3^{2 + \log_5 7 - \log_5 7} = 9$.
A: 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, $$\dfrac{7^{\log_5 15} + 3^{2+\log_5 7}}{7^{\log_5 3}} = 7 + 9\biggl(\dfrac{3^{\log_5 7}}{7^{\log_5 3}}\biggr) = 7 + 9\biggl(\dfrac{x}{y}\biggr) = 7+9(1) = 16.$$
A: $$\frac{7^{ \log_{5} 15    }+3^{2+\log_{5}7}}{7^{\log_{5}3}}=\frac{7^{ 1+\log_53    }+7^{\left(2+\log_{5}7\right)\log_73}}{7^{\log_{5}3}}=7+\frac{7^{\log_79+\log_53}}{7^{\log_53}}=7+9=16.$$
