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Properties of logarithms Quiz 2

mathematics

 Calculate $\dfrac{1}{{{{\log }_2}n}} + \dfrac{1}{{{{\log }_3}n}} + ................. + \dfrac{1}{{{{\log }_{42}}n}} = $ ?

 Use logarithm function,  $\dfrac{1}{{{{\log }_a}b}} = {\log _b}a$ and ${\log _b}m + {\log _b}n = {\log _b}m \times n$ 

 $ \dfrac{1}{{{{\log }_2}n}} + \dfrac{1}{{{{\log }_3}n}} + ................. + \dfrac{1}{{{{\log }_{42}}n}} \\   = {\log _n}2 + {\log _n}3 + ........................... + {\log _n}42 \\   = {\log _n}(2.3.................42) \\   = {\log _n}(42!) \\   = \dfrac{1}{{{{\log }_{42!}}n}}$

 $Use the formula ${\log _b}m - {\log _b}n = \log _b\dfrac{m}{n}$

 ${\log _5}20 - {\log _5}4 = {\log _5}\dfrac{{20}}{4} \\ \Rightarrow {\log _5}\dfrac{{20}}{4} = {\log _5}5 = 1$ 

 Calculate if $x,y,z$ are In G.P. and ${p^x} = {q^y} = {r^z}$

  Take log of given equation and use logarithm function, $\dfrac{{\log a}}{{\log b}} = {\log _b}a$ 

${p^x} = {q^y} = {r^z} \\ x\log p = y\log q = z\log r \\ \Rightarrow \dfrac{y}{x} = \dfrac{z}{y} = \dfrac{{\log p}}{{\log q}} = \dfrac{{\log q}}{{\log r}} \\ \Rightarrow {\log _q}p = {\log _r}q $

 Find ${\log _a}b \times {\log _b}c \times {\log _c}a = $

 Use logarithm function, ${\log _a}b = \dfrac{{{{\log }_m}a}}{{{{\log }_m}b}}$ 

${\log _a}b \times {\log _b}c \times {\log _c}a = \dfrac{{{{\log }_m}b}}{{{{\log }_m}a}} \times \dfrac{{{{\log }_m}c}}{{{{\log }_m}b}} \times \dfrac{{{{\log }_m}a}}{{{{\log }_m}c}}$ 

 If ${\left( {5x} \right)^{\ln 5}} = {\left( {3y} \right)^{\ln 3}}\,and\,{3^{\ln x}} = {5^{\ln y}}$ than $x = $ ?

 Take $\ln $ on both sides of both given equations and solve for values for two equations. $ \\ $ Use logarithm function $ \\ n{\log _b}m = {\log _b}{m^n}$ 

$  \ln {\left( {5x} \right)^{\ln 5}} = \ln {\left( {3y} \right)^{\ln 3}} \\   \Rightarrow \ln 5(\ln 5 + \ln x) = \ln 3(\ln 3 + \ln y).................(1) \\  \ln {(\,3)^{\ln x}} = \ln {(5)^{\ln y}} \\   \Rightarrow \left( {\ln x} \right)\left( {\ln 3} \right) = (\ln y)(\ln 5).............................(2) \\  {\text{using }}(1){\text{ in }}(2){\text{ }} \\   \Rightarrow \ln 5(\ln 5 + \ln x) = \ln 3\left( {\ln 3 + \dfrac{{(\ln x)(\ln 3)}}{{\ln 5}}} \right) \\    \text{On simplifying, we get} \\ \ln ^2\left(5\right)\ln \left(x\right)=\ln ^2\left(3\right)\ln \left(5\right)+\ln ^2\left(3\right)\ln \left(x\right)-\ln ^3\left(5\right) \\ \ln ^2\left(5\right)\ln \left(x\right)-\ln ^2\left(3\right)\ln \left(x\right)=\ln ^2\left(3\right)\ln \left(5\right)-\ln ^3\left(5\right) \\ \ln \left(x\right)\left(\ln ^2\left(5\right)-\ln ^2\left(3\right)\right)=\ln ^2\left(3\right)\ln \left(5\right)-\ln ^3\left(5\right) \\ \ln \left(x\right)=-\ln \left(5\right) \\ x=\dfrac{1}{5} $ 