Free Study Material

NCERT Solutions

NCERT Books

Revision Notes

Practice Problems

This is one of our many quizzes on Simplifying logarithmic expressions. Click on this to access more such competitive quizzes on Exponents and Logarithms.

Simplifying logarithmic expressions Quiz 2

mathematics

 Find the value of $x$ where ${\log _4}x - {\log _4}11 = {\log _2}6?$

 To solve the question, use logarithm identities $\log a - \log b = \log \dfrac{a}{b}$, ${\log _{{b^c}}}a = \dfrac{1}{c}{\log _b}a$ and $b{\log _x}a = {\log _x}{a^b}$.

Given ${\log _4}x - {\log _4}11 = {\log _2}6$ $ \\ $ Now using identity $\log a - \log b = \log \dfrac{a}{b}$ we get: $ \\ {\log _4}\dfrac{x}{{11}} = {\log _2}6 \\   \Rightarrow {\log _{{2^2}}}\dfrac{x}{{11}} = {\log _2}6 \\ $ W.K.T ${\log _{{b^c}}}a = \dfrac{1}{c}{\log _b}a$: $ \\ {\log _{{2^2}}}\dfrac{x}{{11}} = {\log _2}6 \\   \Rightarrow \dfrac{1}{2}{\log _2}\dfrac{x}{{11}} = {\log _2}6 \\   \Rightarrow \log_2 \dfrac{x}{{11}} = 2{\log _2}6 \\ $ Using identity $b\log a = \log {a^b}$ we get:$ \\  \log_2 \dfrac{x}{{11}} = {\log _2}{6^2} \\   \Rightarrow \dfrac{x}{{11}} = {6^2} \\   \Rightarrow x = 11 \times 36 = 396 \\  $

 Find the value of $x$ if $8 + {\log _2}(2x) = 12$.

 To solve the question, use  $   {\log _b}a = x \\   \Rightarrow a = {b^x} \\  $ And $\log ab = \log a + \log b$.

We have  $ 8 + {\log _2}(2x) = 12 \\   \Rightarrow {\log _2}(2x) = 4 \\  $ Now use $\log ab = \log a + \log b$ we get: $  {\log _2}2 + {\log _2}x = 4 \\   \Rightarrow 1 + {\log _2}x = 4 \\   \Rightarrow {\log _2}x = 3 \\  $ W.K.T $ {\log _a}b = x \\   \Rightarrow b = {a^x} \\  $ Using logarithmic identity to solve we get: $  x = {2^3} \\   \Rightarrow x = 8 \\  $

 Find the value of ‘x’ satisfying ${\log _{10}}({20^x} + 7x - 49) = x(2 - {\log _{10}}5)$.

 First simplify write $1$ as ${\log _{10}}10$ and then use the logarithm identity $b{\log _x}a = {\log _x}{a^b}$ as well as $\log a - \log b = \log \dfrac{a}{b}$.

 Given ${\log _{10}}({20^x} + 7x - 49) = x(2 - {\log _{10}}5)$ Now, write $2$ as ${\log _{10}}100$ i.e. $ \\ {\log _{10}}({20^x} + 7x - 49) = x({\log _{10}}100 - {\log _{10}}5) \\ $ Using $\log a - \log b = \log \dfrac{a}{b}$ and then $b{\log _x}a = {\log _x}{a^b}$ we get: $ \\ {\log _{10}}({20^x} + 7x - 49) = x({\log _{10}}20) \\   \Rightarrow {\log _{10}}({20^x} + 7x - 49) = {\log _{10}}{20^x} \\   \Rightarrow {20^x} + 7x - 49 = {20^x} \\   \Rightarrow 7x = 49 \\   \Rightarrow x = 7 \\  $

 Find the value of $x$ where $x = {\log _2}{\log _2}{\log _2}{\log _5}{625^4}$. 

 Solve using logarithm identities $b{\log _x}a = {\log _x}{a^b}$ and ${\log _a}a = 1$

Given:  $ x = {\log _2}{\log _2}{\log _2}{\log _5}{625^4} \\   \Rightarrow x = {\log _2}{\log _2}{\log _2}{\log _5}{5^{4 \times }}^4 \\  $ Using $b{\log _x}a = {\log _x}{a^b}$ and ${\log _a}a = 1$, because $\log 1 = 0$ Now using $\log (a \times b) = \log a + \log b$ we get: $ \\  x = {\log _2}{\log _2}{\log _2}16 \\   \Rightarrow x = {\log _2}{\log _2}{\log _2}{2^4} \\   \Rightarrow x = {\log _2}{\log _2}4 \\   \Rightarrow x = {\log _2}{\log _2}{2^2} \\   \Rightarrow x = {\log _2}2 \\   \Rightarrow x = 1 \\  $

 If ${x^2} - {y^2} = {z^2}$ then find the value of $\dfrac{1}{{{{\log }_{x - z}}y}} + \dfrac{1}{{{{\log }_{x + z}}y}}$

 Solve using ${\log _a}b = \dfrac{1}{{{{\log }_b}a}}$and $\log ab = \log a + \log b$ then use ${x^2} - {y^2} = {z^2}$ to get the answer.

Given $\dfrac{1}{{{{\log }_{x - z}}y}} + \dfrac{1}{{{{\log }_{x + z}}y}}$ $ \\ $ Using logarithm identity ${\log _a}b = \dfrac{1}{{{{\log }_b}a}}$ we get: $ \\ \dfrac{1}{{{{\log }_{x - z}}y}} + \dfrac{1}{{{{\log }_{x + z}}y}} \\   \Rightarrow {\log _y}(x - z) + {\log _y}(x + z) \\  $ Using logarithm identity $\log ab = \log a + \log b$we get: $ \\ {\log _y}[(x - z) \times (x + z)] \\   \Rightarrow {\log _y}({x^2} - {z^2})..........(i) \\  $ Because $(x - z) \times (x + z) = {x^2} - {z^2}$ $ \\ $ We have ${x^2} - {y^2} = {z^2} \Rightarrow {x^2} - {z^2} = {y^2}$ $ \\ $ Using this in $(i)$ we get:$ \\ {\log _y}({x^2} - {z^2}) \\   \Rightarrow {\log _y}{y^2} \\  $ Using $b{\log _x}a = {\log _x}{a^b}$ we get: $ \\ {\log _y}{y^2} = 2{\log _y}y = 2$ Because ${\log _y}y = 1$