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solve for 4*2^5x=21 .48 .42 .08 .31
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\[4\times2^{5x}=21\] Is this correct?
\[2^{5x}=\frac{ 21 }{ 4}\] \[5x \log2= \log (\frac{ 21 }{ 4})\] Solve for x....
so it would be .42?
@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..
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ok i think thats what it is
Hmmm - I suggest you try the calculation again
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