Solve $6^{2x}-9^x=27$

Question

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 :)

Answer 1

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

$$ S=\{1\}$$

Answer 2

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