0
$\begingroup$

Question: $3^{x^2-5} - 9^{x-1} = 0$

$3^{x^2 -5} - 9^{x-1} = 0$

$3^{x^2 -5} = 9^{x-1}$

if $x = 3$

$3^{9-5} = 9^2$

$3^4 = 9^2$

$81 = 81$

and $x = 3$

Is this the right way to solve it?

$\endgroup$
  • 1
    $\begingroup$ Do you know that's the only solution? $\endgroup$ – David Mitra Feb 6 '14 at 18:32
  • $\begingroup$ I didn't get you, sorry! :( Can you explain? @DavidMitra $\endgroup$ – Kiara Feb 6 '14 at 18:35
  • $\begingroup$ You found one solution. There may be others. There might not be others. You need to determine which of the two cases holds and prove it. $\endgroup$ – David Mitra Feb 6 '14 at 18:36
  • $\begingroup$ Yeah, you're right...I have two answers in my book and I got only one. :( $\endgroup$ – Kiara Feb 6 '14 at 18:37
4
$\begingroup$

HINT:

Using Exponent law, $\displaystyle(a^x)^y=a^{xy}$

We have, $$3^{x^2-5}=9^{x-1}=(3^2)^{x-1}$$

$$\implies 3^{x^2-5}=3^{2(x-1)}$$

Now if $\displaystyle a^x=a^y\implies a^{x-y}=1$ (assuming $a$ to be non-zero finite number)

either $a=1$

or $a=-1,x-y$ is even

or $x=y$ for real $a$

$\endgroup$
2
$\begingroup$

That approach certainly works, but you missed a root $x=-1$. A more general approach to the problem is noting that $9 = 3^2$ and that $(x^a)^b = x^{ab}$. Then we have $$3^{x^2-5} - 9^{x-1} = 3^{x^2-5} - (3^2)^{x-1} = 3^{x^2-5} - 3^{2x-2} = 0 \implies 2x-2 =x^2 -5$$

We can just solve for $x$ now: $$2x-2 = x^2 - 5 \implies x^2-2x-3 = (x-3)(x+1) = 0 \implies x=-1 \text{ or } x=3$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.