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My friend asked me this question and he also sent me his solution and just wanted to check if its correct or not. Can you guys check it. What he did was,

$6^{2x}-9^x=6^2-9$

then he equated the terms with the same bases, i.e

$6^{2x}=6^2 and -9^x=-9$

in this way $x = 1$.

I have a feeling that this isn't correct but also cant think of counter example as well. Thank you for your help :)

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    $\begingroup$ You should typeset the question using MathJax so that the equation is clear. Here is a tutorial: math.meta.stackexchange.com/questions/5020/… $\endgroup$
    – Dave
    May 12, 2020 at 2:55
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    $\begingroup$ It isn't a correct method to solve equations in general but it worked this time. $6^2-9$ is indeed equal to $27$ $\endgroup$
    – Alexander Gruber
    May 12, 2020 at 2:57
  • $\begingroup$ Try equating terms with the same bases but with a 27 on the right instead of a 9 (it won't work). The problem is this method gives you two equations in one variable, and so in general they won't have a common solution. $\endgroup$
    – user208649
    May 12, 2020 at 3:01
  • $\begingroup$ in fact, I think $x=1$ is the only solution $\endgroup$ May 12, 2020 at 3:01

2 Answers 2

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It is equivalent to $$9^x4^x-9^x=27$$ which is $$9^x(4^x-1)=27$$ where $f : x \to 9^x(4^x-1)$ is stricly increasing on $\mathbb{R}^+$ hence the solution is unique and

$$ S=\{1\}$$

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  • $\begingroup$ This is a neat approach $\endgroup$
    – imranfat
    May 12, 2020 at 4:30
  • $\begingroup$ Thank you for the help! $\endgroup$ May 13, 2020 at 3:47
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After doing some steps you can see that (4^x)-1=3^(3-2x) Now you can see for the value of x>1 LHS is always integer but RHS is not and when x <0 then LHS is negative but RHS positive now put x= 1 it gives the solution..x has one solution because when you put x=1/2 the mother function is less than 27

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