Question where the exponent is the square root of a log

Question

$4^{\sqrt{\log_4(5)}} -5^{\sqrt{\log_5(4)}}$

The answer is 0 but I do not understand why. I assume you can group the two terms somehow. The question would be much easier if it weren't for the square roots

Answer 1

Follow the steps below,

$$4^{\sqrt{\log_4 5}} =5^{\sqrt{\log_4 5 }}$$

Take $\ln()$ on both sides,

$$ \sqrt{\log_4 5}\ln 4 = \sqrt{\log_5 4} \ln 5$$

Square both sides and use $\log_a b=\ln b/\ln a$,

$$ (\ln 5/ \ln 4)\ln^2 4 = (\ln4/ \ln 5) \ln^2 5$$

$$ \ln 5 \ln 4= \ln4 \ln 5 $$

Answer 2

$4^{\sqrt{\log_4 5}} = k > 4$ (because $\log_4 5 > 1$)

So $5 = 4^{\log_4 5} = (4^{\sqrt{\log_4 5}})^{\sqrt{\log_4 5}} = k^{\sqrt{\log_4 5}}$

Meanwhile $(5^{\sqrt{\log_5 4}})^{\sqrt{\log_4 5}}=5^{\sqrt{\log_54 \cdot \log_4 5}} = 5^1 = 5= k^{\sqrt{\log_4 5}}$.

So $ (4^{\sqrt{\log_4 5}})^{\sqrt{\log_4 5}}=(5^{\sqrt{\log_5 4}})^{\sqrt{\log_4 5}}$

$x^b$ is an invertible function for $x\in \mathbb R$ and $b>0; b\ne 1$.

So $4^{\sqrt{\log_4 5}} = 5^{\sqrt{\log_5 4}}$

Answer 3

Because $$4^{\sqrt{\log_45}}=e^{\ln4\sqrt{\frac{\ln5}{\ln4}}}=e^{\ln5\sqrt{\frac{\ln4}{\ln5}}}=5^{\sqrt{\log_54}}.$$