<s>[INST] How many real solutions does this equation have ? $4^x\times9^\frac{1}{x} + 9^x\times4^\frac{1}{x} + 6^{x+\frac{1}{x}} = 108$ How many real solutions does this equation have ?
$4^x\times9^\frac{1}{x} + 9^x\times4^\frac{1}{x} + 6^{x+\frac{1}{x}} = 108$
I am pretty sure it has something to do with prime factorization of 6, so I wrote
$(2^x\times3^\frac{1}{x})^2 + (3^x\times2^\frac{1}{x})^2 + (2^x\times3^\frac{1}{x})\times(3^x\times2^\frac{1}{x}) = 2^2\times3^3$
If i let $a=(2^x\times3^\frac{1}{x})$ and $b=(3^x\times2^\frac{1}{x})$ the equation will look like this
$a^2+b^2+ab=2^2\times3^3$ which reminds me of $a^3-b^3=(a-b)(a^2+b^2+ab)$.
Unfortunately, I don't know what to do after this, maybe this is not even the right path to solving this?
Any tips ?
Thanks in advance !
 [/INST] You obtained the form $$a^2+b^2+ab=108$$
Dividing both sides with $ab$, we obtain the following equation:
$$(\frac 23)^{x-\frac 1x}+\frac {1}{(\frac 23)^{x-\frac 1x}}+1=\frac {108}{6^{x+\frac 1x}}$$
Note that, from the original equation, it can be easily shown that there can never be any negative solutions,  since all three terms are $<1$. I shall leave this to you. For $x>0$, we have:
$$x+\frac 1x\geq 2$$
This means that $$\frac {108}{6^{x+\frac 1x}}\leq \frac {108}{36}\leq3$$
However, from AM-GM on LHS, we get that LHS$\geq 3$. Hence only solution is $x=1$, when both sides equal $3$.
</s>