Mathematics · Quantitative Aptitude

Logarithms

246 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 differencial calculus - differenciability and methods of differnciation differentiation by substitution methods of differentiation derivative of a function

Differential coefficient of $\log\ \sin x$ is :

  1. $\cos x$
  2. $\tan x$
  3. $\text{cosec} \,x$
  4. $\cot x$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

We have,

$y=\log \sin x$

On differentiating w.r.t $x$, we get
$\dfrac{dy}{dx}=\dfrac{d(\log \sin x)}{dx}$
$\dfrac{dy}{dx}=\dfrac{1}{\sin x}\times \cos x$
$\dfrac{dy}{dx}=\dfrac{\cos x}{\sin x}$
$\dfrac{dy}{dx}=\cot x$

Hence, this is the answer.

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.

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

Which of the following statements is not correct?

  1. $log _{10} 10 = 1$
  2. $log (2+ 3) = log (2 \times 3)$
  3. $log _{10} 1 = 0$
  4. $log (1 + 2 + 3) = log 1 + log 2 + log 3$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

(a) Since $log _a a = 1,$ so $log _{10}  10 = 1$
(b) $log (2 + 3) log 5$ and $log (2 \times 3) = log 6 = log 2 + log 3$
$\therefore log(2 + 3) \neq log (2 \times 3)$
(c) Since, $log _a  1 = 0$, so, $log _{10} 1 = 0$.
(d) $log(1 + 2 + 3) = log 6 = log (1 \times 2 \times 3) = log 1 + log 2 + log 3$