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This is the second of our many quizzes on Log rules. Click on this to access more such competitive quizzes on Number System.

Log rules Quiz 2

mathematics

   Find  ${{\log }_{2}}(20)-{{\log }_{2}}(10)$.

  Apply basic log rules to solve the expressions separately and then subtract them.

  $ \text{We have the log rule, lo}{{\text{g}}_{b}}x=y,x={{b}^{y}} \\  \therefore {{\log }_{2}}(20)={{2}^{4.32}}=20=4.32 \\  {{\log }_{2}}(10)={{2}^{3.32}}=10=3.32 \\  \therefore {{\log }_{2}}(20)-{{\log }_{2}}(10)=4.32-3.32=1 \\ $

 If   ${{e}^{4\ln (2x)}}=256$. Then find $x$.

 Using the basic log rules simplify ${{e}^{4\ln (2x)}}$ and then solve the expressions to get the answer.

  $ \text{Applying log rules,} \\  {{e}^{4\ln (2x)}}=256 \\  \Rightarrow {{e}^{\ln {{(2x)}^{4}}}}=256 \\  \Rightarrow {{(2x)}^{4}}=256 \\  \Rightarrow x=2  $

  Given ${{\log }_{3}}81-{{\log }_{3}}(x-4)=2$. What is $x$?

 Using the basic log rule ${{\log }_{b}}(x)-{{\log }_{b}}(y)={{\log }_{b}}\left( \dfrac{x}{y} \right)$ and simplify the given expression to get the required value.

 $ \text{Applying log rule, i}\text{.e}\text{., }{{\log }_{b}}(x)-{{\log }_{b}}(y)={{\log }_{b}}\left( \dfrac{x}{y} \right),\text{ we get} \\  {{\log }_{3}}81-{{\log }_{3}}(x-4)=2 \\  \Rightarrow {{\log }_{3}}\left( \dfrac{81}{x-4} \right)=2 \\  \Rightarrow \left( \dfrac{81}{x-4} \right)={{3}^{2}}=9 \\  \Rightarrow 9x-36=81 \\  \therefore x=13 $

 Simply $\log \left( 10.\dfrac{3\sqrt{x}}{20}.4{{x}^{2}} \right)$.

  Using the basic log rule ${{\log }_{b}}(xy)={{\log }_{b}}\left( x \right)+{{\log }_{b}}\left( y \right)$ and simplify the given expression to get the required value.

 $ \text{Applying log rule, i}\text{.e}\text{., }{{\log }_{b}}(xy)={{\log }_{b}}\left( x \right)+{{\log }_{b}}\left( y \right),\text{ we get} \\  \log \left( 10.\dfrac{3\sqrt{x}}{20}.4{{x}^{2}} \right)=\log (10)+\log \left( 3 \right)+\log \left( \dfrac{\sqrt{x}}{20} \right)+\log \left( {{x}^{2}} \right)+\log \left( {{x}^{2}} \right) \\  \Rightarrow \log \left( 10.\dfrac{3\sqrt{x}}{20}.4{{x}^{2}} \right)=1+\log \left( 3 \right)+\log \left( \dfrac{\sqrt{x}}{20} \right)+\log \left( {{x}^{2}} \right)+\log \left( {{x}^{2}} \right) \\  \Rightarrow \log \left( 10.\dfrac{3\sqrt{x}}{20}.4{{x}^{2}} \right)=1+\log \left( 3 \right)+\log \left( \sqrt{x} \right)-\log \left( 20 \right)+\log (4)+2\log \left( x \right) $

  What is the square root of the difference between the coefficients of terms with $\log $ and the constant term in the simplification of  $\log \left( \dfrac{7{{x}^{2}}}{4} \right)+\log \left( 3{{x}^{3}}\sqrt{x} \right)-\log \left( 10 \right)+\dfrac{3}{4}\ln \left( e \right)$

 Using the basic log rule ${{\log }_{b}}(xy)={{\log }_{b}}\left( x \right)+{{\log }_{b}}\left( y \right),\log \left( 10 \right)=1,\ln \left( e \right)=1$, simplify the given expression. Now find the difference between the coefficient of the log term and constant and then take square root.

Applying log rule, i.e., $ \\ {{\log }_{b}}(xy)={{\log }_{b}}\left( x \right)+{{\log }_{b}}\left( y \right),\log \left( 10 \right)=1,\ln \left( e \right)=1,\text{ we get} \\  \log \left( \dfrac{7{{x}^{2}}}{4} \right)+\log \left( 3{{x}^{3}}\sqrt{x} \right)-\log \left( 10 \right)+\dfrac{3}{4}\ln \left( e \right)=\log \left( \dfrac{7{{x}^{2}}}{4}\left( 3{{x}^{3}}\sqrt{x} \right) \right)-1+\dfrac{3}{4} \\  \Rightarrow \log \left( \dfrac{7{{x}^{2}}}{4} \right)+\log \left( 3{{x}^{3}}\sqrt{x} \right)-\log \left( 10 \right)+\dfrac{3}{4}\ln \left( e \right)=\log \left( \dfrac{21}{4}\left( {{x}^{5}}\sqrt{x} \right) \right)-\dfrac{1}{4} \\ $ So, the square root of the difference between the coefficients of terms with $\log $ and the constant term will be, $ \sqrt{1-\dfrac{1}{4}}=\dfrac{\sqrt{3}}{2} $