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Mathematics 14 Online
OpenStudy (anonymous):

which of these are equivalent to 16^x? select all those that apply. A. (4*4)^x B. 4^x*4^x C. 4^2*4^x D. 4*4^x E. 4*4^2x F. 4^2x

OpenStudy (anonymous):

I have A so far

OpenStudy (anonymous):

You'll have to use the following facts: \[16=4\times4\] \[\large\left(a^b\right)^c=a^{bc}\]

OpenStudy (anonymous):

A is right. What else can you get?

OpenStudy (anonymous):

well... I was thinking B???

OpenStudy (anonymous):

Right! Because \((4^x)\times(4^x)=(4^x)^2=4^{2\times x}=(4^2)^x=16^x.\)

OpenStudy (anonymous):

There's at least one more

OpenStudy (anonymous):

so... just A & B?

OpenStudy (anonymous):

oh

OpenStudy (anonymous):

idk

OpenStudy (anonymous):

Look at the series of equalities I wrote earlier:\[4^{2\times x}=~?\]

OpenStudy (anonymous):

4^2+x

OpenStudy (anonymous):

right? ..........

OpenStudy (anonymous):

A. \((4*4)^x=16^x\) B. \(4^x*4^x=4^{2x}\) C. \(4^2*4^x=4^{2+x}\) D. \(4*4^x=4^{x+1}\) E. \(4*4^2x=4^{2x+1}\) F. \(4^{2x}=4^{2x}\)

OpenStudy (anonymous):

so that's the 3rd one?

OpenStudy (anonymous):

so it'd be A, b, & f?

OpenStudy (anonymous):

no

OpenStudy (anonymous):

right? it'd just be A, B, & F

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