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

Logarithm Quiz 2

mathematics

 Which of the following represents the graph of $f\left( 2x+4 \right)$ if $f\left( x \right)=\log x$

 Calculate the answer by using the rules of transformation of graphs.

  $\text{From our knowledge of logarithms, we can say graph of }f\left( x \right)=\log x\text{ is} \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/33d1bec1-d6fa-4bae-9505-701374164a792584623046076729997.png"/>$\\ \therefore\text{ The graph of }f\left( 2x \right)\text{ will shrink horizontally by }\dfrac{1}{2}, \\ \text{ because graph of }f\left(ax \right)\text{ shrinks by }\dfrac{1}{a},\text{ if }\left| a \right| > 1.$ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/83f19986-d78a-4e17-9f86-27961d8c268f6272840401782742808.png"/> $\\ \therefore \text{Graph of }f\left( 2x+4 \right)\text{ will shift left, because graph of }f\left( x+a \right)\text{ shifts left by “a”,} \\ \text{If }\left| a \right|>1. \\ \text{ Hence, we get the below figure after modifications. } \\ $<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/25025a6d-9310-42ea-b5ea-ddbcf1ec8df04020891359134790486.png"/>

 If ${{\log }_{10}}6+\log \left( 2x-1 \right)=\log 4+\log \left( 6x+2 \right)$, then

 Use the property $\log a+\log b=\log \left( a\times b \right)$ and then take antilog.

  $\text{Using different logarithmic properties, we get} \\ {{\log }_{10}}6+\log \left( 2x-1 \right)=\log 4+\log \left( 6x+2 \right) \\ \Rightarrow \log \left[ 6\left( 2x-1 \right) \right]=\log \left[ 4\left( 6x+2 \right) \right] \\ \Rightarrow 12x-6=24x+8 \\ \Rightarrow 12x=-14 \\ \Rightarrow x=-\dfrac{7}{6}$

 The roots of ${{\log }_{4}}\left( {{4}^{2x}}-{{4}^{x}}+1 \right)=x$ is

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

 $\text{Using the property of logarithms, we get} \\ {{\log }_{4}}\left( {{4}^{2x}}-{{4}^{x}}+1 \right)=x \\ \Rightarrow {{4}^{2x}}-{{4}^{x}}+1={{4}^{x}} \\ \Rightarrow {{\left( {{4}^{x}}-1 \right)}^{2}}=0 \\ \Rightarrow {{4}^{x}}=1 \\ \Rightarrow x=0$

 If $y=\log \dfrac{a}{b}+\log \dfrac{c}{a}+\log \dfrac{b}{c}$, then evaluate the value of $y$.

 Solve this question by simply using $\log a+\log b+\log c=\log \left( a\times b\times c \right)$.

 $\text{Using the basic concepts of logarithm, we get} \\ y=\log \dfrac{a}{b}+\log \dfrac{c}{a}+\log \dfrac{b}{c} \\ y=\log \left( \dfrac{a}{b}\times \dfrac{c}{a}\times \dfrac{b}{c} \right) \\ \Rightarrow y=\log 1 \\ \Rightarrow y=0$

 Find the number of values of x for the following equation:<br>${{\log }_{5}}x+{{\log }_{4}}x=0$

 Use the property that ${{\log }_{b}}a=\dfrac{1}{{{\log }_{a}}b}$ and $\log a+\log b=\log \left( a\times b \right)$.

 $\text{Let us assume x=1, we get} \\ {{\log }_{5}}\left( 1 \right)+{{\log }_{4}}\left( 1 \right)=0 \\ \Rightarrow 0=0 \\ \text{Let us consider, }x\ne 1,\text{ we get} \\ {{\log }_{5}}x+{{\log }_{4}}x=0 \\ \Rightarrow \dfrac{1}{{{\log }_{5}}x}+\dfrac{1}{{{\log }_{4}}x}=0 \\ \Rightarrow {{\log }_{x}}5+{{\log }_{x}}4=0 \\ \Rightarrow {{\log }_{x}}\left( 5\cdot 4 \right)=0 \\ \Rightarrow 20=1,\text{ which is not possible.} \\ \text{Therefore, the only possible value of “x” is 1.} \\ \therefore \text{There exists only one value of “x”.}$