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Mathematics 22 Online
OpenStudy (anonymous):

What is the solution of the following equation?

OpenStudy (anonymous):

\[2^{2x} = 8^{x + 1}\]

OpenStudy (phi):

re-write 8 as \(2^3\)

OpenStudy (anonymous):

So then it would be \[2^{2x} = 2^{3x + 1}\] , or...?

OpenStudy (phi):

almost. use the idea \[ \left(a^b\right)^c = a^{bc}\] multiply the exponents. in your case \[ 2^{2x} = (2^3)^{(x + 1)} \]

OpenStudy (phi):

notice that 3*(x+1) is not 3x+1

OpenStudy (phi):

you distribute the 3: 3*x + 3*1

OpenStudy (anonymous):

Ohh okay! I see what I did wrong

OpenStudy (phi):

\[ 2^{2x} = 2^{3x + 3} \] common sense should tell you that 2 to some power 2x is the same as 2 to some power 3x+3 means the exponents must be the same number: 2x must equal 3x+3 2x = 3x+3 solve for x

OpenStudy (anonymous):

Okay! I can take it from here (: Again, thank you so much!

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