# Solve the following equation.

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?

• Do you know that's the only solution? – David Mitra Feb 6 '14 at 18:32
• I didn't get you, sorry! :( Can you explain? @DavidMitra – Kiara Feb 6 '14 at 18:35
• 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. – David Mitra Feb 6 '14 at 18:36
• Yeah, you're right...I have two answers in my book and I got only one. :( – Kiara Feb 6 '14 at 18:37

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 $a$

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: $$2x-2 = x^2 - 5 \implies x^2-2x-3 = (x-3)(x+1) = 0 \implies x=-1 \text{ or } x=3$$