Given that $9^{2x} = 27^{x^2 - 5}$. Find the possible values of $x$.

Question

Given that $9^{2x} = 27^{x^2 - 5}$. Find the possible values of $x$.

I don't know how to approach this question.

Answer 1

Hints:

1. $9=3^2$
2. $27=3^3$
3. $(a^b)^c = a^{bc}$
4. $a^x = a^y$ if and only if $x=y$.

Answer 2

Hints$$9 = 3^2 \implies 9^z = 3^{2z} $$ $$27 = 3^3 \implies 27^y = 3^{3y}$$

Now compare both sides.

Answer 3

Using that $$3^{4x}=3^{3x^2-15}$$ we get the equation $$4x=3x^2-15$$

Answer 4

Hint:

We have \begin{align} 9^{2x} = 27^{x^2 - 5} &\iff \exp(\ln(9)2x) = \exp(\ln(27)(x^2-5))\\ &\iff \frac{\ln(9)}{\ln(27)}2x -x^2 + 5=0 \end{align}

Answer 5

As $9=3^2, 9^{2x}=(3^2)^{2x}=3^{4x}$

Similarly, $27^{x^2-5}=3^{3x^2-15}$

As $3^{4x}\ne0,$

$$\implies1=\dfrac{3^{3x^2-15}}{3^{4x}}=3^{3x^2-4x-15}$$

Now like Find all real numbers $x$ for which $\frac{8^x+27^x}{12^x+18^x}=\frac76$,

if $\displaystyle u^m=1,$

either $\displaystyle m=0 $

or $\displaystyle u=1$

or $\displaystyle u=-1,m$ is even

But here $u=3\ne\pm1$