Find $x^\sqrt{2}+y^{\sqrt{2}}$ if $(\log_2 x)^4 + (\log_3 y)^4 + 8 = 8 (\log_2 x)(\log_3 y)$ and $x, y > 1$.

Question

Find $x^\sqrt{2}+y^{\sqrt{2}}$ if $(\log_2 x)^4 + (\log_3 y)^4 + 8 = 8(\log_2 x)(\log_3 y)$ and $x, y > 1$.

Because I do not like the logarithms, I will substitute $a = \log_2 x$ and $b = \log_3 y$. Because $x, y > 1$, we know that $a, b > 0$. We want to find $x^{\sqrt{2}}+y^{\sqrt{2}}$, so we try to get $x, y$ in terms of $a, b$. We get that $x = 2^a$ and $y = 3^b$, so what we want to find is $2^{\sqrt{2}a}+3^{\sqrt{2}b}$.

Using the equation that we are given, we get that $$a^4 + b^4 + 8 = 8ab$$ I have no idea how to continue from here, and $a, b > 0$ must be there for a reason, but I don't know how to use it.

Answer 1

Suppose $(a,b)$ is a solution to $a^4+b^4+8=8ab$ with $b=ka$. Then $$a^4(1+k^4)-8ka^2+8=0$$ As a quadratic in $a^2$, this has discriminant $$\Delta=64k^2-32(1+k^4)=-32(k^2-1)^2$$ which is negative unless $k=\pm1$. So the only solutions have $b=\pm a$. Actually since you want $a,b>0$, they are equal. Now you just need to solve $$2a^4+8=8a^2$$ $$a^4- 4a^2+4=0$$ $$(a^2- 2)^2=0$$

And you can conclude what $a$ must be, and then what $b$ is, and then...

Answer 2

This is a bit of a cheating way, but here it goes:

We still use the substitution $a = \log_2 x$ and $b = \log_3 y$. Then, $a^4+b^4+8=8ab$. We see a $a^4$ and $b^4$, and we see a $ab$ on the other side, so we think of using AM-GM for 4 variables.

We split 8 into $4+4$ so that there are 4 terms. Applying AM-GM to the left side, $a^4 + b^4 + 4 + 4 \geq 4\sqrt[4]{(a^4)(b^4)(4)(4)}$. The right side of the inequality is equal to $8ab$, so we know that for $a, b$ to satisfy the problem condition, the equality condition of AM-GM must be satisfied. We have that $a^4=b^4=4$, or $a=b=\sqrt{2}$. We plug this into what we want to get the answer of $13$.