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This is the second of our many quizzes on solving logarithmic equations. Click on this to access more such competitive quizzes on Logarithm.

Solving Logarithmic Equations Quiz 2

mathematics

 The value of $x$ satisfying given equation is ${{\log }_{10x}}\left( {{x}^{2}}+25 \right)=1$

 Use the property, ${{\log }_{a}}b=c\Leftrightarrow b={{a}^{c}}$ and then by square the equation.

  $\begin{align} & \text{Using basic properties of logarithmic equations, we get}\\ & \Rightarrow {{x}^{2}}+25={{\left( 10x \right)}^{1}}\\ & \Rightarrow {{\left( x-5 \right)}^{2}}=0 \\ & \Rightarrow x=5\end{align}$

 If ${{\log }_{4}}\left( 7x-2 \right)={{\log }_{4}}\left( 5 \right)$, then $x$ is

 Take antilogs on both sides of the equation.

  $\begin{align} & \text{Using basic property of logarithmic equations, we get}\\ & {{\log }_{4}}\left( 7x-2 \right)={{\log }_{4}}\left( 5 \right)\\ & \Rightarrow 7x-2=5\\ & \Rightarrow x=1\end{align}$

 For which equation, does ${{\log }_{3}}\left( 2x \right)+{{\log }_{3}}\left( 2x+6 \right)={{\log }_{3}}\left( 7 \right)$ is a solution?

 Use the property ${{\log }_{a}}b+{{\log }_{a}}c={{\log }_{a}}bc$.

 $\begin{align} & \text{Using basic property of logarithmic equation, we get}\\ & {{\log }_{3}}\left( 2x \right)+{{\log }_{3}}\left( 2x+6 \right)={{\log }_{3}}\left( 7 \right)\\ &\Rightarrow {{\log }_{3}}\left( 2x \right)\cdot \left( 2x+6 \right)={{\log }_{3}}\left( 7 \right)\\ &\Rightarrow 4{{x}^{2}}+12x-7=0\\ &\Rightarrow x=-\dfrac{3}{2},\dfrac{1}{2}\end{align}$

 Which of the following equation has the value of $x$ as $\dfrac{-b+\sqrt{{{b}^{2}}+4ac}}{2a},\dfrac{-b-\sqrt{{{b}^{2}}+4ac}}{2a}$?

 Solve it either by taking antilogs or using property ${{\log }_{a}}\left( a \right)=1$.

 $\begin{align} & \text{Use properties of logarithmic equation, we get} \\ & {{\log }_{c}}\left( a{{x}^{2}}+bx \right)={{\log }_{c}}c\\ & \Rightarrow a{{x}^{2}}+bx=c\\ &\Rightarrow x=\dfrac{-b+\sqrt{{{b}^{2}}+4ac}}{2a},\dfrac{-b-\sqrt{{{b}^{2}}+4ac}}{2a}\end{align}$

 If $y={{\log }_{4}}x$ then ${{\log }_{4}}\left( 8{{x}^{2}} \right)$ is equal to

 Solve by using property ${{\log }_{a}}b+{{\log }_{a}}c={{\log }_{a}}bc$ and ${{\log }_{a}}b=c\Leftrightarrow b={{a}^{c}}$.

 $\begin{align} & \text{Use properties of logarithmic equation, we get} \\ & {{\log }_{4}}\left( 8{{x}^{2}} \right)\\ &\Rightarrow {{\log }_{4}}8+{{\log }_{4}}{{x}^{2}}\\ &\Rightarrow \dfrac{3}{2}+2{{\log }_{4}}x\\ &\Rightarrow \dfrac{3}{2}+2y\end{align}$