SOLUTION: solve for x: 2log2x-log21=log29

Question 436590: solve for x: 2log2x-log21=log29
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
solve for x: 2log2x-log21=log29
..
2log2x-log21=log29
2log2x-log21-log29=0
2log2x-(log21+log29)=0
log((2x)^2)/(21*29)=0
convert to exponential form: (base raised to log of number=number) In this case base=10, log of the number=0, and the number=((2x)^2)/(21*29)
10^0=4x^2/609=1
4x^2=609
x^2=609/4
x=+-sqrt(609/4)=12.33896
x=12.33896
x=-12.33896 (reject, 2x>0)
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