Question

Algebra 2 help would be greatly appreciated!

Answer 1

\[3^{2x}-15(3^{x})+56=0\]

Answer 2

Hey so what's it asking

Answer 3

Solve for x? :O

Answer 4

Yeah solve for x

Answer 5

substitute 3^x as a and solve the equation as a quadratic equation

Answer 6

do you need more explanation ?i mean can do it on your own

Answer 7

@science00000 ??

Answer 8

can i multiply the -15 and 3^x ???

Answer 9

No but you can simplify it -5(3^x+1)

Answer 10

\[-5(3^{x+1})\]

Answer 11

don't seem to be quite understanding this :/

Answer 12

You have to know your exponent rules, 5 x 3 = 15, so we have 5(3)(3^x) you can only do this with like bases so we have \[3^{x+1}\]

Answer 13

So here is the general rule \[x^mx^n = x^{m+n}\]

Answer 14

You can factor this I suppose

Answer 15

Yes, factoring should work

Answer 16

factoring you should end up with \[(3^x-8)(3^x-7)=0\] as we are treating 3^x as a, also suggested above

Answer 17

ohh this makes a little more sense now!

Answer 18

-7 x - 8 = 56
-7+-8=15

Answer 19

\[\large 3^{2x}-15(3^{x})+56=0\]
\[\large (\color{red}{3^{x}})^2-15(\color{red}{3^{x}})+56=0\]

let \(t=\color{red}{3^{x}}\)

\[\large (\color{red}t{})^2-15(\color{red}{t})+56=0\]
you knw how to handle a quadratic

Answer 20

-15*

Answer 21

Yes!! Thanks everyone for helping! Really appreciate it.

Answer 22

You can solve for x now, it involves logarithms, you know how to do that right? \[3^x - 7 = 0\] set it to 0 and solve for x for both

Answer 23

Yeah of course. Thanks again everyone!

Answer 24

Ok, yw :)