0

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 :)

Alapan Das
  • 936
Kritesh Dhakal
  • 580
  • 2
    You should typeset the question using MathJax so that the equation is clear. Here is a tutorial: https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – Dave May 12 '20 at 02:55
  • 6
    It isn't a correct method to solve equations in general but it worked this time. $6^2-9$ is indeed equal to $27$ – Alexander Gruber May 12 '20 at 02:57
  • 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. –  May 12 '20 at 03:01
  • in fact, I think $x=1$ is the only solution – J. W. Tanner May 12 '20 at 03:01

2 Answers2

5

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

$$ S=\{1\}$$

EDX
  • 1,797
0

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

Sayan
  • 39