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Mathematics 15 Online
OpenStudy (anonymous):

solve for 4*2^5x=21 .48 .42 .08 .31

OpenStudy (mrnood):

\[4\times2^{5x}=21\] Is this correct?

OpenStudy (mrnood):

\[2^{5x}=\frac{ 21 }{ 4}\] \[5x \log2= \log (\frac{ 21 }{ 4})\] Solve for x....

OpenStudy (anonymous):

so it would be .42?

OpenStudy (anonymous):

@MrNood

OpenStudy (mrnood):

I don't really care about the answer and have not calculated it.# Have you rearranged the above equation for x=??? If you are happy with that and the answer is what you get then fine. If you have not done the calculation then ask if there is a point you are stuck on..

OpenStudy (anonymous):

ok i think thats what it is

OpenStudy (mrnood):

Hmmm - I suggest you try the calculation again

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