Question: Context: I am preparing for jee and this is the question I have encountered on text book, I do have solution but I am curious on why should we divide this equation with $4^{\log\_{10}(x)}$, and we need to find $x$.

$$
4(4^{\log\_{10}{x}}) - 6^{\log\_{10}{x}}-18(9^{\log\_{10}{x}}) = 0
$$

dividing o.b.s with $4^{\log\_{10}(x)}$  

$$
4-(\frac{6}{4})^{\log\_{10}(x)} - 18(\frac{9}{4})^{\log\_{10}(x)} = 0
$$

$$
4-(\frac{3}{2})^{\log\_{10}(x)} - 18((\frac{3}{2})^2)^{\log\_{10}(x)} = 0
$$

put $t = \frac{3}{2}^{\log\_{10}(x)}$
$$
4-t-18t^2 = 0
$$

**Again the question is how to solve or know when to divide with what to find solution not only this question but is there kinda procedure when you get struck?**

I know its kinda complicated to explain but how to find breakthrough in logarithmic problems like these ones?
Edit: any other way to find $x$ other than traditional text book method?

Answer: 
I have some other solution which can you save you from thinking about such substitutions

$4(4^{\log\_{10}{x}}) - 6^{\log\_{10}{x}}-18(9^{\log\_{10}{x}}) = 0$

$4{(2^{\log\_{10}{x}})}^2 - 2^{\log\_{10}{x}}.3^{\log\_{10}{x}}-18{(3^{\log\_{10}{x}})}^2 = 0$

$4x^2-xy-18y^2=0$ where $x=2^{\log\_{10}{x}}$ and $y=3^{\log\_{10}{x}}$. Now factorization gives you

$(4x-9y)(x+2y)=0$ Now $x+2y>0$, therefore

$4x=9y$

$2^{2+\log\_{10}{x}}=3^{2+\log\_{10}{x}}$, which is possible when ${2+\log\_{10}{x}}=0$

Hence $x=\frac{1}{100}$
It's of the pattern $m.a^{2x}+n.(ab)^{x}+r.b^{2x}=0(a,b\gt0)$

To solve it, you divide by $b^{2x}$ and put $(\frac ab)^x=t$ for $t\gt0$

You will get a quadratic in $t$.