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This is the second of our many quizzes on Common and natural logarithms. Click on this to access more such competitive quizzes on Number System.

Common and natural logarithms Quiz 2

mathematics

 Solve $\log [4 + \ln (e) + {\log _3}(2x + 2)] = 1$ and mark the correct option. 

 Apply the property : $  {\log _a}b = x \\   \Rightarrow x = {a^b} \\  $

  Simplifying and then applying the property : $ \\ {\log _a}b = x \\   \Rightarrow x = {a^b} \\  $.  We know that $\log 10 = 1$and $\ln (e) = 1$. Substituting these values we get, $ \\ \log [4 + \ln (e) + {\log _3}(2x + 2)] = 1 \\   \Rightarrow \log [4 + 1 + {\log _3}(2x + 2)] = \log 10 \\  $ Taking antilog both sides we get, $ \\  \log [4 + 1 + {\log _3}(2x + 2)] = \log 10 \\   \Rightarrow 5 + {\log _3}(2x + 2) = 10 \\   \Rightarrow {\log _3}(2x + 2) = 5 \\   \Rightarrow 2x + 2 = {3^5} \\   \Rightarrow 2x = 241 \\   \Rightarrow x = 120.5 \\  $

 Solve ${\log _4}64 - {\log _4}(x - 5) = 4$ and choose the correct answer.

 Use the property  ${\log _b}x - {\log _b}y = {\log _b}(\dfrac{x}{y})$. 

  Applying the property ${\log _b}x - {\log _b}y = {\log _b}(\dfrac{x}{y})$ we get, $ \\   {\log _4}64 - {\log _4}(x - 5) = 4 \\   \Rightarrow {\log _4}(\dfrac{{64}}{{x - 5}}) = 4 \\   \Rightarrow \dfrac{{64}}{{x - 5}} = {4^4} = 256 \\   \Rightarrow 64 = 256x - 1280 \\   \Rightarrow x = 5.25 \\  $

 Find the square root of the difference between the coefficient of the terms with $\log $ and constants terms in simplification of $\log (\dfrac{{9{x^2}}}{5}) + \log (4{x^3}\sqrt x ) - \log (10) + \dfrac{4}{3}\ln (e)$.

 Apply the property ${\log _b}(xy) = {\log _b}x + {\log _b}y$.

We know ${\log _b}(xy) = {\log _b}x + {\log _b}y$,$\log (10) = 1$ and $\ln (e) = 1$. Applying these properties., we get, $ \\   \log (\dfrac{{9{x^2}}}{5}) + \log (4{x^3}\sqrt x ) - \log (10) + \dfrac{4}{3}\ln (e) \\   \Rightarrow \log (\dfrac{{36}}{5}{x^5}\sqrt x ) - 1 + \dfrac{4}{3} \\   \Rightarrow \log (\dfrac{{36}}{5}{x^5}\sqrt x ) + \dfrac{1}{3} \\  $ Now the difference between the coefficient of the $\log $ and constants is: $ \\ \sqrt {1 - \dfrac{1}{3}}  = \sqrt {\dfrac{2}{3}} $

 Solve ${\log _{20x}}({x^2} + 4) = 1$ and select the  correct answer.

 Apply the property $  {\log _a}b = c \\   \Rightarrow b = {a^c} \\   $ and use the formula $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ to get the value of $x$.

 Applying the property : $ \\ {\log _a}b = c \\   \Rightarrow b = {a^c} \\  $ We get, $ \\  {\log _{20x}}({x^2} + 4) = 1 \\   \Rightarrow {x^2} + 4 = {(20x)^1} \\   \Rightarrow {x^2} - 20x + 4 = 0 \\  $ Applying the formula $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$. We get, $ \\   x = \dfrac{{ - ( - 20) \pm \sqrt {400 - 4(1)(4)} }}{{2(1)}} \\   \Rightarrow x = \dfrac{{20 \pm \sqrt {384} }}{2} \\   \Rightarrow x = 19.79,0.205 \\  $

 Select the correct simplification option for  ${\log _6}x + {\log _8}x + \ln (e) = 0$

 Use the property ${\log _b}a = \dfrac{1}{{{{\log }_a}b}}$.

 Applying the property ${\log _b}a = \dfrac{1}{{{{\log }_a}b}}$ and $\log a + \log b = \log (ab)$. We get, $ \\   {\log _6}x + {\log _8}x + \ln (e) = 0 \\   \Rightarrow \dfrac{1}{{{{\log }_6}x}} + \dfrac{1}{{{{\log }_8}x}} + 1 = 0 \\   \Rightarrow {\log _x}6 + {\log _x}8 + 1 = 0 \\   \Rightarrow {\log _x}(6 \times 8) + 1 = 0 \\   \Rightarrow {\log _x}48 =  - 1 \\  $