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

megan0430

New member
Joined
Aug 23, 2006
Messages
30
Find "x" so that "a" would not equal 0 and a^x^2 * a^6x = a^-8

PLEASE HELP!
THANK YOU SO MUCH
 

stapel

Super Moderator
Staff member
Joined
Feb 4, 2004
Messages
15,943
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
Joined
Dec 15, 2005
Messages
2,396
\(\displaystyle \L (a^x)^2 \times a^{6x} =\)

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

\(\displaystyle \L a^{8x}\)

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

soroban

Elite Member
Joined
Jan 28, 2005
Messages
5,588
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: \(\displaystyle \L\:x\:=\:-2,\;-4\)

 
Top