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How would you solve log_2 (x/y) = 3?

 Mar 5, 2017
 #1
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log_2 (x/y) = 3?

 

y = x/8 = (log(x/y))/(log(2)) = 3 = log(x/y) = log(8) = log((8 y)/x) = 0

 Mar 5, 2017
 #2
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Using the change of base formula:

 

log2(x/y)=log10(x/y)log102=3 log10(x/y)=3(log102) log10(x/y)=log10(23) x/y=23 x/y=8

 

So x = 8y    and     y = x/8

 

I don't know what Guest did, but I'm pretty sure y can't be zero because that would make a zero in the denominator and that would mean that it doesn't equal 3, it would just not be a true statement.

 Mar 5, 2017

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