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How do you know which one is bigger between $5^\sqrt{3}$ and $4^\sqrt{5}$?

For my method I used in $2^\sqrt{3}$ and $3^\sqrt{2}$ I put both numbers in the function $f(x)=x^\sqrt{3}$ so $f(2^\sqrt{3})$ become $2^3$ which is equal to 8 and $f(3^\sqrt{2})$ become $3^\sqrt{6}$ and $3^\sqrt{6}>3^\sqrt{2}$ which is equal to 9 and 9>8 so $3^\sqrt{2}>2^\sqrt{3}$

But for this problem I don't know what I should multiply or is there any method besides that? Please kindly help. Thank you

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    $\begingroup$ Did you mean $3^{\sqrt6}>3^{\sqrt4}=9$? $\endgroup$ Oct 19, 2019 at 1:53

2 Answers 2

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Take both to the power of $\sqrt{5}$

$\left(5^\sqrt{3}\right)^\sqrt{5}=5^{\sqrt{15}}<5^{\sqrt{16}}=5^4=625$

$\left(4^\sqrt{5}\right)^\sqrt{5}=4^5=1024$

so we have that $4^\sqrt{5}>5^\sqrt{3}$

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We make a power of $\sqrt{3}$ both expression:

$\left(5^\sqrt{3}\right)^\sqrt{3}=5^3=125$

$\left(4^\sqrt{5}\right)^\sqrt{3}=4^\sqrt{15}>4^{3.5}=4^3\times 2=128>125$

Therefore, $4^\sqrt{5}>5^\sqrt{3}$

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