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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?

Gary
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Rambal heart remo
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  • I'm aspiring for 2022 – Rambal heart remo Jun 04 '21 at 15:31
  • ok consider deleting these comments as they are not related to question – Lalit Tolani Jun 04 '21 at 15:33
  • $a=2^{\log_{10} x}, b=3^{\log_{10} x} $ and you have $4a^2-ab-18b^2=0$ from here you can solve $a$ in terms of $b$ or notice that dividing by $a^2$ or $b^2$ gives a quadratic in $b/a, a/b$ respectively. – kingW3 Jun 04 '21 at 15:47
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    I good piece of advice to approach such problems is: eliminate what disturbs you! Dividing by $4^{\log_{10}(x)}$ does this: it reduces the equation by one ugly term. Of course, it is chosen in order to get the quadratic equation. Without "seeing" this, one would simply divide successively by all terms: $4^{\log_{10}(x)}$ , $6^{\log_{10}(x)}$ , $9^{\log_{10}(x)}$ and see where it leads to. However, eliminating what sucks most is always a good idea, e.g. for integrations. – Marius S.L. Jun 04 '21 at 15:55

2 Answers2

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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 $x=\frac{1}{100}$

Lalit Tolani
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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 $t$.

aarbee
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