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Logarithmic equations Quiz 2

mathematics

 Find the value of: ${\log _{\sqrt[3]{2}}}\sqrt[3]{{16}}$ 

 Use these formulae to simplify: ${\log _{{a^q}}}{m^p} = \dfrac{p}{q}{\log _a}m$ and ${\log _a}a = 1$ 

$ \Rightarrow {\log _{\sqrt[3]{2}}}\sqrt[3]{{16}} \\ \\ $ We can rewrite the expression as, $ \\ \Rightarrow {\log _{{2^{\frac{1}{3}}}}}{16^{\frac{1}{3}}} \\ \Rightarrow \dfrac{{\dfrac{1}{3}}}{{\dfrac{1}{3}}}{\log _2}16 \\ \Rightarrow {\log _2}16 \\ \Rightarrow {\log _2}{2^4} \\ \Rightarrow 4{\log _2}2 \\ \Rightarrow 4$

 If $\dfrac{{\log x}}{2} = \dfrac{{\log y}}{3} = \dfrac{{\log z}}{5}$, then find $xy \text{ ( in terms of } z)$

 Use this formula to add logs with same base: ${\log _a}m + {\log _a}n = {\log _a}(mn)$

 Let, $ \Rightarrow \dfrac{{\log x}}{2} = \dfrac{{\log y}}{3} = \dfrac{{\log z}}{5} = k \\ \Rightarrow \log x = 2k \\ \Rightarrow \log y = 3k \\ \Rightarrow \log z = 5k \\ \Rightarrow \log x + \log y = 2k + 3k \\ \Rightarrow \log xy = 5k \\ \Rightarrow \log xy = \log z \\ $ Hence, $ \Rightarrow xy = z$

 To simplify use the formula: ${\log _{{a^q}}}{m^p} = \dfrac{p}{q}{\log _a}m$

 $ \Rightarrow {\log _2}{\log _2}256 \\ \Rightarrow {\log _2}{\log _2}{2^8} \\ \Rightarrow {\log _2}8 \\ \Rightarrow {\log _2}{2^3} \\ \Rightarrow 3$ 

  Simplify: ${\log _{16}}{12^{{{\log }_5}[{{\log }_2}3 \times {{\log }_3}2]}}$ 

 Simplify the expression using the formula: ${\log _a}1 = 0$ and ${\log _a}b = \dfrac{1}{{{{\log }_b}a}}$ 

 $ \Rightarrow {\log _{16}}{12^{{{\log }_5}[{{\log }_2}3 \times {{\log }_3}2]}} \\ \Rightarrow {\log _{16}}{12^{{{\log }_5}\left[{{\log }_2}3 \times \frac{1}{{{{\log }_3}2}}\right]}} \\ \Rightarrow {\log _{16}}{12^{{{\log }_5}1}} \\ \Rightarrow {\log _{16}}{12^0} \\ \Rightarrow {\log _{16}}1 \\ \Rightarrow 0$

 Which of the following is true, if ${\log _{16}}({\log _{12}}x) < 0$ 

 If $x = {\log _a}N$, then N is a positive number and a is a positive number other than $1$ .

 $ \Rightarrow {\log _{16}}({\log _{12}}x) < 0 \\ \Rightarrow {\log _{12}}x < 1 \\ \Rightarrow x < 12\;\;and\;\;x > 1 \\ \Rightarrow 1 < x < 12$ 