# Which is bigger 5^(√3) and 4^(√5)?

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

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

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

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