Question #12 Please help

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26-01-2018
Mathematics

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Question #12 Please help

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jdoe0001 jdoe0001 · Answer 1 · 2018-01-26T23:22:09+01:00

[tex]\bf \textit{logarithm of factors}\\\\ log_a(xy)\implies log_a(x)+log_a(y) \\\\\\ \textit{Logarithm of rationals}\\\\ log_a\left( \frac{x}{y}\right)\implies log_a(x)-log_a(y) \\\\\\ \textit{Logarithm of exponentials}\\\\ log_a\left( x^ b \right)\implies b\cdot log_a(x)\\\\ -------------------------------\\\\ [/tex]

[tex]\bf a)\\\\ log_9\left( \sqrt[3]{y^2+y} \right)\implies log_9\left[ (y^2+y)^{\frac{1}{3}} \right]\implies \cfrac{1}{3}log_9(y^2+y) \\\\\\ \cfrac{1}{3}log_9[y(y+1)]\implies \cfrac{1}{3}\left[~log_9(y)~+~log_9(y+1) ~ \right] \\\\\\ \cfrac{1}{3}log_9(y)~~+~~\cfrac{1}{3}log_9(y+1)[/tex]

[tex]\bf b)\\\\ log_4\left( \frac{a^2b}{\sqrt{c}} \right)\implies log_4(a^2b)~-~log_4(\sqrt{c})\implies log_4(a^2b)~-~log_4(c^{\frac{1}{2}}) \\\\\\\ [~log_4(a^2)~+log_4(b)~]~~-~~\cfrac{1}{2}log_4(c)\\\\\\ log_4(a^2)+log_4(b)-\cfrac{1}{2}log_4(c)[/tex]