Question

Solve following equation. See the attachment.

Answer 1

\[3*2^{\log_{x}(3x-2) }+2*3^{\log_{x}(3x-2) }-5*6^{\log_{x^2}(3x-2) } =0\]

Answer 2

Latex is not working!!!

Answer 3

!!

Answer 4

lol my brain isn't working, latex is just fine ! ahhahaha....

Answer 5

log x^2 a = log x a/ log x x^2 = 1/2 log x a

Answer 6

my keyboad is also not working LOL

Answer 7

try drawing...

Answer 8

3 * 2^y + 2* 3^y - 5*6^(y/2) = 0

Answer 9

@ErkoT where do you come up with these problems?
they're so random (and tricky)

Answer 10

3 * 2^y + 2* 3^y - 5*2^(y/2) * 3^(y/2) = 0

Answer 11

There's an exam coming soon and we have to solve completely random tests at home.

Answer 12

3 * 2^y - 3*2^(y/2) * 3^(y/2) + 2* 3^y - 2*2^(y/2) * 3^(y/2) = 0
3(2^y - 2^(y/2) * 3^(y/2)) + 2(3^y - 2^(y/2) * 3^(y/2)) = 0
32^(y/2)(2^(y/2) - 3^(y/2)) + 2*3^(y/2)(3^(y/2) - 2^(y/2)) = 0

looks much simpler now.

Answer 13

(2^(y/2) - 3^(y/2))(3*2^(y/2) - 2*3^(y/2)) = 0

put back log x (3x-2) = y ...

Answer 14

doesn't look easy at all ...

Answer 15

straining my eyes trying to read exX's answer lol\[3 * 2^y - 3*2^{y/2} 3^{y/2} + 2* 3^y - 2*2^{y/2} * 3^{y/2} = 0\]\[
3(2^y - 2^{y/2} * 3^{y/2}) + 2(3^y - 2^{y/2}* 3^{y/2}) = 0\]\[
3*2^{y/2}(2^{y/2} - 3^{y/2}) + 2*3^{y/2}(3^{y/2} - 2^{y/2}) = 0\]

Answer 16

Can we use this property