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use the laws of loagrithms to simplify ln(x^2y^4z^7)
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= ln(x^2) + ln(y^4) + ln(z^7) = 2ln(x) + 4ln(y) + 7ln(z)
\[\log _{2}(x ^{2}\sqrt{2x-3})\]
= log2(x^2) + log2((2x - 3)^(1/2)) = 2log2(x) + 1/2*log2(2x - 3)
how did you get (1/2)?
the square root of m can be written as m^(1/2) proof: m^(1/2) * m^(1/2) = m^(1/2 + 1/2) = m^(1) = m hence (m^(1/2))^2 = m hence m^(1/2) = sqrt(m)
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one last question
\[\log _{2}(\sqrt[3]{3x-5}/\sqrt[4]{2x+1})\]
= log2((3x - 5)^(1/3)) - log2((2x + 1)^(1/4)) = 1/3*log2(3x - 5) - 1/4*log2(2x + 1) do you get it now
remember that the n'th root of y can be written as y^(1/n)
yeaa ido thanks for yur help:)
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\[\ln{x^2y^4z^7}=\ln{x^2}+\ln{y^4}+\ln{z^7}=2\ln{|x|}+4\ln{|y|}+7\ln{z}\]\[x\neq0,y\neq0,z>0\]
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