Lost Logarithm

Let $x = \log_{5/4}{4}$ and $y = \log_{5/4}{5}$ .

Then $y = \log_{5/4}({\dfrac{5}{4}.4}) = \log_{5/4}{\dfrac{5}{4}} + \log_{5/4}{4} = 1 + \log_{5/4}{4} = 1 + x$

$\therefore$ The expression = $\dfrac{4^x}{5^y} = \dfrac{4^x}{5^{1+x}} = (\dfrac{5}{4})^{-x}.\dfrac{1}{5} = (\dfrac{5}{4})^{-\log_{5/4}{4}}.\dfrac{1}{5} = (\dfrac{5}{4})^{\log_{5/4}{\dfrac{1}{4}}}.\dfrac{1}{5} = \dfrac{1}{4}.\dfrac{1}{5}= \dfrac{1}{20}$

Note that $\large{a^{\log_{a}{x}} = x}$ . So the expression $= 0.05$

Let $Q$ define the given the expression (where $x=4$ and $y=5$ ).

$Q=\frac { { x }^{ \log _{ \frac { y }{ x } }{ (x )} } }{ { y }^{ \log _{ \frac { y }{ x } }{ (y)} } }= \frac { { x }^{ \frac { ln{ (x) } }{ ln{( \frac { y }{ x} }) } } }{ { y }^{ \frac { ln{ (y) } }{ ln{( \frac { y }{ x } ) } } } }$

Apply the natural logarithm: $ln{ (Q })=ln{ ({ x }^{ \frac { ln({ x) } }{ ln{ (\frac { y }{ x } ) } } }) }-ln{ ({ y }^{ \frac { ln{ (y) } }{ ln{ (\frac { y }{ x }) } } }) }=\frac { ln{ (x) } }{ ln{ (\frac { y }{ x } ) } } ln{ (x) } -\frac { ln(y) }{ \ln { (\frac { y }{ x } ) } } \ln { (y) }$

Simplify the equation: $\ln{(Q)} = \frac { { ln(x) }^{ 2 }-{ ln(y) }^{ 2 } }{ ln({ \frac { y }{ x } ) } } =\frac { \ln { (xy)\cdot \ln { (\frac { x }{ y } ) } } }{ \ln { (\frac { y }{ x } ) } } =-\ln {( xy) }$

$\ln { (Q) =-\ln { (xy) \Longrightarrow Q= \frac { 1 }{ xy } } }$

Substitute $x=4$ and $y=5$ : ${Q=\frac { 1 }{ 4\cdot 5 }=\frac { 1 }{ 20 } }$