Evaluate without calculator or logarithm help

On my previous blog post(Evaluate $12^{\frac{1-(x+y)}{2(1-y)}}$), we're asked, without using the calculator and the help from logarithm, evaluate $\large12^\frac{1-(x+y)}{2(1-y)}$ provided $3=60^x$ and $5=60^y$.

Let do this problem as we're told, where we could not borrow help from calculator nor logarithms.

If we multiply the two given exponential equations, we get:

$3\cdot 5=60^x\cdot 60^y$
   
$3\cdot 5=60^{x+y}$

$3\cdot 5=5^{x+y}\cdot 12^{x+y}$

$3=5^{x+y-1}\cdot 12^{x+y}(*)$

$\scriptsize \log_{24}(48)+\log_{12} (54)>4$

Prove that $\left(\log_{24}(48) \right)^2+ \left(\log_{12}(54) \right)^2 >4$.

For any logarithms problem (be it solving a logarithmic equation or proving a logarithmic inequality), one would usually use the change of base formula to rewrite the two logarithms terms so that they have the same base. Also, one would be sorely tempted to rewrite the LHS of the inequality by breaking down the figures 12, 24, 48 and 52 into the product of prime numbers:

[MATH]\color{yellow}\bbox[5px,green]{12=2^2\cdot 3}[/MATH], [MATH]\color{yellow}\bbox[5px,purple]{24=2^3\cdot 3}[/MATH], [MATH]\color{yellow}\bbox[5px,blue]{48=2^4\cdot 3}[/MATH] and [MATH]\color{yellow}\bbox[5px,orange]{54=2\cdot 3 ^3}[/MATH]