# [MOVED] Exponent Prob: Find x so that a is not zero and...

#### megan0430

##### New member
Find "x" so that "a" would not equal 0 and a^x^2 * a^6x = a^-8

THANK YOU SO MUCH

#### stapel

##### Super Moderator
Staff member
Your formatting is ambiguous. Do you mean "(a^x)^2 times (a^6) times x", "a^(x^2) times (a^6) times x", or something else?

What have you tried? How far have you gotten?

Thank you.

Eliz.

#### skeeter

##### Senior Member
$$\displaystyle \L (a^x)^2 \times a^{6x} =$$

$$\displaystyle \L a^{2x} \times a^{6x} =$$

$$\displaystyle \L a^{8x}$$

since $$\displaystyle \L a^{8x} = a^{-8}$$, can you figure out what x is?

#### soroban

##### Elite Member
Re: [MOVED] Exponent Prob: Find x so that a is not zero and.

Hello, Megan!

If I interpreted your notation correctly, I have the solution . . .

$$\displaystyle \text{Find }x:\;\;\L a^{^{x^2}}\cdot a^{^{6x}}\:=\:a^{-8}\;$$$$\displaystyle \text{ for }a\,\neq\,0$$

Multiply both sides by $$\displaystyle a^8:$$$$\displaystyle \L\;\;\left(a^{^{x^2}}\right)\left(a^{^{6x}}\right)\left(a^{^8})\;=\;1$$

$$\displaystyle \;\;$$This becomes: $$\displaystyle \L\:a^{^{(x^{^2}+6x+8)}}\;=\;1\;=\;a^{^0}$$

And we have the quadratic: $$\displaystyle \L\:x^2\,+\,6x\,+\,8 \:=\:0$$

$$\displaystyle \;\;$$which factors: $$\displaystyle \L\x\,+\,2)(x\,+\,4)\;=\;0$$

$$\displaystyle \;\;$$ and has roots: $$\displaystyle \L\:x\:=\:-2,\;-4$$