Question

2^2x-2^x=6

Answer 1

hello joemath!

Answer 2

all yours

Answer 3

let:
\[y=2^x\iff y^2=(2^x)^2=2^{2x}\]Using this substitution gives:
\[y^2-y=6\iff y^2-y-6=0\iff (y-3)(y+2)=0\iff y=3,-2\]
Now replacing y back with 2^x gives:\[y=3\iff 2^x=3\iff x=\log_2(3)\]and\[y=-2\iff 2^x=-2\] which has no solution.

Answer 4

ok is there any other simpler terms because you just lost me

Answer 5

unfortunately no there isnt a simpler way. The only way to solve this is to notice the equation is a quadratic type equation that can be factored.

Answer 6

by hand anyways. im sure a calculator could do it, or maybe wolfram.

Answer 7

wait hoe exactly would i factor this, that same thing you just showed me?

Answer 8

yes. its hard to see how it factors in its original form, but after making the substitution its a little easier to see it.

Answer 9

not to butt in but do you want a decimal rather than
\[\log_2(3)\]?

Answer 10

exact answer only!!!! lolol jk

Answer 11

i just need to solve for x

Answer 12

well that it "exact" but it really begs the question.
you say "what would i raise the number 2 to the power of to get 3?" and i reply
\[\log_2(3)\]which is only a different way of stating the question

Answer 13

btw if you want to look fancy and impress your math teacher you can factor as joemath did but just use
\[2^x\] instead of y. factor as
\[(2^x-3)(2^x+2)=0\]

Answer 14

Not many people can see that right off the bat. thats the purpose of the substitution.

Answer 15

how would that solve for x??

Answer 16

i agree. you solve for
\[2^x=3\] via
\[x=\frac{\ln(3)}{\ln(2)}\] in english "the log of the total divided by the log of the base"

Answer 17

thank you both