Find "x" so that "a" would not equal 0 and a^x^2 * a^6x = a^-8 PLEASE HELP! THANK YOU SO MUCH
M megan0430 New member Joined Aug 23, 2006 Messages 30 Aug 23, 2006 #1 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 16,583 Aug 24, 2006 #2 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.
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 Elite Member Joined Dec 15, 2005 Messages 3,215 Aug 24, 2006 #3 \(\displaystyle \L (a^x)^2 \times a^{6x} =\) \(\displaystyle \L a^{2x} \times a^{6x} =\) \(\displaystyle \L a^{8x}\) since \(\displaystyle \L a^{8x} = a^{-8}\), can you figure out what x is?
\(\displaystyle \L (a^x)^2 \times a^{6x} =\) \(\displaystyle \L a^{2x} \times a^{6x} =\) \(\displaystyle \L a^{8x}\) since \(\displaystyle \L a^{8x} = a^{-8}\), can you figure out what x is?
S soroban Elite Member Joined Jan 28, 2005 Messages 5,586 Aug 25, 2006 #4 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 . . . \(\displaystyle \text{Find }x:\;\;\L a^{^{x^2}}\cdot a^{^{6x}}\:=\:a^{-8}\;\)\(\displaystyle \text{ for }a\,\neq\,0\) Click to expand... 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: \(\displaystyle \L\:x\:=\:-2,\;-4\)
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 . . . \(\displaystyle \text{Find }x:\;\;\L a^{^{x^2}}\cdot a^{^{6x}}\:=\:a^{-8}\;\)\(\displaystyle \text{ for }a\,\neq\,0\) Click to expand... 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: \(\displaystyle \L\:x\:=\:-2,\;-4\)