SOLUTION: Explain why 2^x = 15 cannot always be solved in the same way as 2^x = 16. How would you solve each?

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Question 961264: Explain why 2^x = 15 cannot always be solved in the same way as 2^x = 16. How would you solve each?
Found 2 solutions by stanbon, MathLover1:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
Explain why 2^x = 15 cannot always be solved in the same way as 2^x = 16. How would you solve each?
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16 is a power of 2,
So 2^x = 2^4
x = 4
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15 is hard to evaluate as a power of 2
so 2^x = 15
x*log(2) = log(15)
x = log(15)/log(2)
x = 3.0746
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Cheers,
Stan H.
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Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!


2%5Ex+=+15 since 15 cannot be written as exponent with base 2, to solve it we need to take log of both sides
log%282%5Ex%29+=+log%2815%29
x%2Alog%282%29+=+log%283%2A5%29
x%2Alog%282%29+=+log%283%29%2Blog%285%29
x=+%28log%283%29%2Blog%285%29%29%2Flog%282%29+

and
2%5Ex+=+16 since 16 cant be written as exponent with base 2 as 2%5E4, we will have
2%5Ex+=+2%5E4...since base same , we have
x=4