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Mathematics 14 Online
OpenStudy (anonymous):

solve for x: 4^x-2^(x+1)=15

OpenStudy (anonymous):

\[4^{x}-2^{x+1}=15\]

OpenStudy (anonymous):

\[ 4^x=2^{2x} \]

OpenStudy (anonymous):

But then what do I do?

OpenStudy (anonymous):

\[ 2^{x+1}=2\cdot 2^x \]

OpenStudy (anonymous):

You can factor that out at least.

OpenStudy (anonymous):

How?

OpenStudy (anonymous):

Try \(u=2^x\) maybe

OpenStudy (anonymous):

I am completely lost

OpenStudy (anonymous):

\[ 4^{x}-2^{x+1}=15 \]\[ (2^x)^2-2(2^x)=15 \]\(u=2^x,x=\log_2(u)\)\[ u^2-2u=5 \]

OpenStudy (anonymous):

you don't know how to find the values of \(u\)?

OpenStudy (anonymous):

No, this is like an intro to logs unit

OpenStudy (anonymous):

do you know how to solve a quadratic equation? @iamnotasalad

OpenStudy (anonymous):

Yes

OpenStudy (anonymous):

Solve for \(u\) then....

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