Question 47490
 
log2 x+log2(x+3)-log2x^2
Using the log laws you can rewrite this expression as:
log[x*(x+3)/x^2
=log2 [(x^2+3x)/x^2]
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solve x: -5=log2x
2^(-5)=x
x=1/32
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e^5x+1=40 solve 
e^(5x+1)=40
Take the natural log of both sides to get:
5x+1 = ln40
5x+1 = 3.68887945...
5x=2.68887945...
x=0.53777589...
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7log+log2 32 
This problem is garbled
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if f(x)=log2x find f(16)
f(16)=log2 16 = 4
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Cheers,
Stan H.