If $\log_22016=a$, $\log_32016=b$, $\log_72016=c$, evaluate $\frac{1}{2a}+\frac{1}{5b}+\frac{1}{10c}$
My work: I've added the fractions to be $$\frac{5bc+2ac+ab}{10abc}$$ and substituted $a, b, c$ with the change of base formula, but I can't get it to simplify. When you plug in the expression into a calculator you get $\frac{1}{10}$, but I'm not sure what log properties will get me there without a calculator.
Hint:
Since $\log_22016=a$, so $\frac{1}{a}=\log_{2016}2$. This means \begin{align*} \frac{1}{2a}+\frac{1}{5b}+\frac{1}{10c}& =\frac{1}{2}\log_{2016}2+\frac{1}{5} \log_{2016}3 + \frac{1}{10}\log_{2016}{7}\\[10pt] & =\log_{2016}\sqrt{2}+\log_{2016}\sqrt[5]{3}+\log_{2016}\sqrt[10]{7}\\[10pt] & =\log_{2016}\sqrt{2}\cdot \sqrt[5]{3}\cdot\sqrt[10]{7} \end{align*}
\begin{eqnarray*} \frac{1}{10}( 5 \log_{2016} 2 +2 \log_{2016} 3 + \log_{2016} 7 ) \end{eqnarray*} and note that $2016=2^5 \times 3^2 \times 7$.
Note: If $\log_m n = k$ then $m^k = n$ and (assuming $m > 0$ then $m = n^{\frac 1k}$ so $\log_n m = \frac 1k$.
Or in other words $\frac 1{\log_m n} = \log_n m$
So
$\frac 1{2a} + \frac 1{5b} + \frac 1{10c} = $
$\frac 12 \log_{2016}2 + \frac 15 \log_{2016} 3 + \frac 1{10} \log_{2016} 7=$
$\log_{2016}2^{\frac 12}*3^{\frac 15}*7^{\frac 1{10}}$
Now the trick is to realize that if $m = 2^{\frac 12}*3^{\frac 15}*7^{\frac 1{10}}$ then $m^{10} = 2^5*3^2*7 = 2016$ so $m = 2016^{\frac 1{10}}$.
So $\log_{2016}2^{\frac 12}*3^{\frac 15}*7^{\frac 1{10}}= \log_{2016} 2016^{\frac 1{10}} = \frac 1{10}$
Let us see how the expression has come into being.
$$7^c=2016\iff7^{c-1}=3^22^5\iff5\log2+2\log3+(1-c)\log7=0$$
Similarly, form the other two equations in $\log2,\log3,\log5$
Using Cramer's Rule, as the right hand is $0(d_1,d_2,d_3$ in the link)
we must have $$\begin{vmatrix} 5 & 2 & 1-c \\ 5 & 2-b & 1 \\ 5-a & 2 & 1 \notag \end{vmatrix}=0\implies abc=5bc+ab+2ca$$
