Lost Logarithm IV

We know that $\dfrac 1{\log_a{b}} = \log_b{a}$

$\begin{aligned} \dfrac 2{\log_4{2000^6}} + \dfrac 3{\log_5{2000^6}} &= 2 \cdot \dfrac{1}{\log_4{2000^6}} + 3\cdot \dfrac {1}{\log_5{2000^6} }\\ &=2{\log_{2000^6}{4}} + 3{\log_{2000^6}{5}} \\ &={\log_{2000^6}{4^2}} + {\log_{2000^6}{5^3}}\\ &={\log_{2000^6}{4^2 \cdot 5^3}}\\ &={\log_{2000^6}{2000}}\\ &= {\frac{1}{6}}.\end{aligned}$

Therefore our answer is $1 + 6 = \boxed{7}$ .

First, $\frac { 2 }{ \log _{ 4 }{ { 2000 }^{ 6 } } } +\frac { 3 }{ \log _{ 5 }{ { 2000 }^{ 6 } } } \\ =\frac { 2 }{ 6\log _{ 4 }{ 2000 } } +\frac { 3 }{ 6\log _{ 5 }{ { 2000 } } } \\ =\frac { 1 }{ 3\log _{ 4 }{ 2000 } } +\frac { 1 }{ 2\log _{ 5 }{ { 2000 } } } \\$

After some things, we get $\\ \frac { 1 }{ 3\left( 2+3\log _{ 4 }{ 5 } \right) } +\frac { 1 }{ 2\left( 3+2\log _{ 5 }{ 4 } \right) } \\$

Let $\log _{ 5 }{ 4 } \\$ be $x$

$\\ \frac { 1 }{ 3\left( 2+\frac { 3 }{ x } \right) } +\frac { 1 }{ 2\left( 3+2x \right) } \\ =\frac { 3\left( 2+\frac { 3 }{ x } \right) +2\left( 3+2x \right) }{ 6\left( 3+2x \right) \left( 2+\frac { 3 }{ x } \right) } \\ =\frac { 6+\frac { 9 }{ x } +6+4x }{ 6\left( 6+\frac { 9 }{ x } +6+4x \right) } \\ =\frac { 1 }{ 6 }$

so the answer is $\boxed{\boxed{\boxed{\boxed{7}}}}$