Question

Solve for x
9^2(x+2) = 27^x-2

Answer 1

\[\large 9^{2(x+2)}=27^{x-2}\] right?

Answer 2

since \(9=3^2\) and \(27=3^3\) you can write this as
\[\large 3^{6(x+2)}=3^{3(x-2)}\] making
\[6(x+2)=3(x-2)\]and that is not too hard to solve

Answer 3

do we have to make the 6 equal to 3?

Answer 4

that would be a miracle, wouldn't it?

Answer 5

ya

Answer 6

oh darn i made a mistake
lets try again, this time i go slow and step by step

Answer 7

ok lol

Answer 8

this is where we start \[\large 9^{2(x+2)}=27^{x-2}\] yes?

Answer 9

yes

Answer 10

then since \(9=3^2\) and \(27=3^3\) we can turn this in to
\[\large \left(3^2\right)^{2(x+2)}=\left(3^3\right)^{x-2}\]

Answer 11

ya true now the bases are the same

Answer 12

then by the laws of exponents, you multiply the exponents and get
\[\large 3^{4(x+2)}=3^{3x-6}\]

Answer 13

since as you said, the bases are the same, now you know that
\[4(x+2)=3x-6\] and that is what you need to solve for \(x\)
you good with that?

Answer 14

just give me a sec to solve it

Answer 15

k

Answer 16

4x +2 = 3x-6
1x=-6-2
1x/1 = -8/1
x=-8

Answer 17

if you start with \[4(x+2)=3x-6\] you have to remove the parentheses on the left using the distributive property
first step should be
\[4x+8=3x-6\]

Answer 18

oh ok the 4x-3x = 1x
-8-6 = -14

Answer 19

that is what i get, yes
so assuming the first equation was right, that answer should be right \(x=-14\)

Answer 20

ok thank you so much

Answer 21

yw