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Given log2 a = x and log4 b = y then log2 (a³/b²) is equal to: a) 3x + y b) 3x + 4y c) 3x - y d) 3x - 4y No idea how to do this one
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b) 3x + 4y ..
how did you get that?
\[\log_2(\frac{a^3}{b^2})=\log_2(a^3)-\log_2(b^2)=3\log_2(a)-2\log_2(b)=3x-2\log_2(b)\] Now that one part we have to think about \[\log_4(b)=y => 4^y=b =>(2^2)^y=b => 2^{2y}=b => \log_2(b)=2y\] So we have \[\log_2(\frac{a^3}{b^2}) =3x-2 (2y)=3x-4y\]
Thank you so much!!
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