Mathematics · Quantitative Aptitude

Logarithms

231 Questions

Logarithms are mathematical operations that determine the exponent required for a base to reach a specific number. This topic tests the application of logarithmic properties, changing bases, and solving complex equations. It is a high-yield topic for quantitative aptitude in competitive exams.

Logarithmic expressionsBase change propertiesSolving log equationsInfinite series logsCharacteristic values

Logarithms Questions

Multiple choice maths set concepts finite and infinite sets types of sets set language

Let S be the set of all values of x such that $log _{2x}(x^{2}+5x+6)<1$ then the sum of all integral value of x in the set S, is

  1. 0

  2. 8

  3. 9

  4. 10
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$ log _{2x}(x^{2}+5x+6)< 1 $

$ \Rightarrow x^{2}+5x+6< 2x^1 $

$ \Rightarrow x^{2}+3x+6< 0 $

But $ x^{2}+3x+6 = 0 $ has no real roots 

$ \therefore S$  is an empty set 

$ \therefore $ sum of all integral values of $ x = 0 $ 
Multiple choice business maths linear programming problems structure of linear programming model linear programming problem operations research

If $x=\log _{2^2}2+\log _{2^3}2^2+\log _{2^4}2^3......+\log _{2^{n+1}}2^n+$, then the minimum value of $x$ will be-

  1. $\left(\dfrac 1{n+1}\right)^{\tfrac 1n}$
  2. $n\left(\dfrac 1{n+1}\right)^{\tfrac 1n}$
  3. $\left(\dfrac n{n+1}\right)^{\tfrac 1n}$

  4. None of the above.
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
given $x={log _{2^2}2+log _{2^3}{2^2}+log _{2^4}{2^3}+...+log _{2^{n+1}}2^n}$

we know that $A.M. \geq G.M.$

$\implies \dfrac{log _{2^2}2+log _{2^3}{2^2}+log _{2^4}{2^3}+...+log _{2^{n+1}}2^n}{n}\geq \sqrt[n]{log _{2^2}2*log _{2^3}{2^2}*log _{2^4}{2^3}*...*log _{2^{n+1}}{2^n}}$

$\implies \dfrac{x}{n}\geq \sqrt[n]{\dfrac{log2}{log2^2}*\dfrac{log2^2}{log2^3}*\dfrac{log2^3}{log2^4}*...*\dfrac{log2^n}{log2^{n+1}}}$

$\implies x\geq n\sqrt[n]{\dfrac{log2}{log2^{n+1}}}$

$\implies x\geq n({\dfrac{log2}{(n+1)log2}})^{\dfrac 1n}$

$\implies x\geq n({\dfrac{1}{n+1}})^{\dfrac 1n}$

therefore the minimum value of $x$ is $ n({\dfrac{1}{n+1}})^{\dfrac 1n}$
Multiple choice physics logarithms introduction to logarithm logarithmic notation exponential and logarithms

If $\log\ (-2x)=2\log\ (x+1)$, then $x$ can be  equal to

  1. $-2+\sqrt {3}$
  2. $-4+2\sqrt {3}$
  3. $-2-\sqrt {3}$
  4. $-4-2\sqrt {3}$
Reveal answer Fill a bubble to check yourself
A,C Correct answer
Explanation

We have,

$ \log \left( -2x \right)=2\log \left( x+1 \right) $

$ \Rightarrow \log \left( -2x \right)=\log {{\left( x+1 \right)}^{2}} $


Comparing both side and we get,

$ -2x={{\left( x+1 \right)}^{2}} $

$ \Rightarrow -2x={{x}^{2}}+1+2x $

$ \Rightarrow {{x}^{2}}+4x+1=0 $


Using quadratic formula and we get,

$ x=\dfrac{-4\pm \sqrt{16-4\times 1\times 1}}{2\times 1} $

$ x=\dfrac{-4\pm \sqrt{12}}{2} $

$ x=\dfrac{-4\pm \sqrt{2\times 2\times 3}}{2} $

$ x=\dfrac{-4\pm 2\sqrt{3}}{2} $

$ x=-2\pm \sqrt{3} $


Hence, this is the answer.

Multiple choice physics logarithms introduction to logarithm logarithmic notation exponential and logarithms

Express the following in logarithmic form$\,\colon$
$81\,=\,3^{4}$

  1. $\log _381\,=\,4$
  2. $\log _981\,=\,2$
  3. $2\log _39\,=\,4$
  4. $4\log _93\,=\,2$
Reveal answer Fill a bubble to check yourself
A,B,C,D Correct answer
Explanation

$y={ a }^{ x }\Rightarrow \log _{ a }{ y } =x\ \therefore 81={ 3 }^{ 4 }\Rightarrow \log _{ 3 }{ 81 } =4$

A is true.
$81=3^{4}=9^{2}$

$\Rightarrow \log _{9}81=\log _{9}9^{2}=2\log _{9} 9=2$
B is true.
$81=3^{4}=9^{2}$
$\Rightarrow \log _{3} 3^{4}=\log _{3}9^{2}=2\log _{3} 9$
C is true.
$81=3^{4}=9^{2}$
$\Rightarrow \log _{9}3^{4}=\log _{9}9^{2}=2\log _{9} 9=2$
$\Rightarrow 4\log _{9} 3=2$
D is true.