Question

.

Answer 1

ok you are thinking that
\[3^{2x}=(3^x)^2\] so if you put
\[z=3^x\] you have a quadratic equation
\[z^2=5z+36\]
\[z^2-5z-36=0\]
\[(z-9)(z+3)=0\]
\[z=9,z=-3\]

Answer 2

now replace z by 3^x get
\[3^x=9,3^x=-3\]

Answer 3

you can forget about
\[3^x=-3\] because no matter what x is,
\[3^x>0\] so it cannot be -3

Answer 4

as for
\[3^x=9\] it is clear that \[3^2=9\] meaning
\[x=2\]

Answer 5

forget the damned logs, they are not important, and i see i made a small mistake, it should be
\[(z-9)(z+4)=0\] but the result is the same

Answer 6

Wow that actually makes so much more sense! Thanks @satellite73 ! The logs were confusing me

Answer 7

you do not need logs to solve
\[3^x=9\] or
\[3^x=27\] you might need them later for more difficult problems

Answer 8

Alright. thanks again @satellite73, you explanation makes so much more sense

Answer 9

@satellite73, how does it go from z2−5z−36=0 to (z−9)(z+3)=0? what happens here?

Answer 10

that was a typo i corrected it
should be
\[z^2-5z-36=(z-9)(z+4)\]

Answer 11

Oh okok. Thank you!