Tag: logarithms

Questions Related to logarithms

$\log V = 2 \log 2 - \log 3 + \log \pi + 3 \log r$ can be expressed as

physics logarithms introduction to logarithm logarithmic notation exponential and logarithms

  1. $V = \dfrac{4}{3} \pi r^{3}$

  2. $ V = \dfrac{2}{3} \pi r^{3}$

  3. $ V = \dfrac{4}{3} \pi r$

  4. $ V = \dfrac{2}{3} \pi r$

Show answer
Correct Option: A
Explanation:
$ \log { V }  = 2\log { 2 } -\log { 3 } +\log { \pi  }  + 3\log { r }$
$ \log V = \log ({ 2 }^{ 2 }\times \pi \times { r }^{ 3 }) - \log { 3 }$
$ \log V = \log \dfrac { 4\pi { r }^{ 3 } }{ 3 } $
Removing log
$V = \dfrac {4}{3} \pi r^3$

Which of the following is true for $\log _25$?

physics logarithms introduction to logarithm logarithmic notation exponential and logarithms

  1. An integer

  2. A rational number

  3. An irrational number

  4. A whole number

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Correct Option: C
Explanation:

Let us assume that $\log _{ 2 }{ 5 } =\frac { p }{ q } $ , where $p,q$ are integers

We have $5=2^{\frac{p}{q}}$
$\Rightarrow 5^{q}=2^{p}$
This suggest that $5$ and $2$ are not mutually prime , But $2$ and $5$ are mutually prime
Therefore $\log _{ 2 }{ 5 } $ is an irrational number

If $\log _{10}(x - 10) = 1$, then value of $x$ is

physics logarithms introduction to logarithm logarithmic notation exponential and logarithms

  1. $10$

  2. $13$

  3. $20$

  4. $26$

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Correct Option: C
Explanation:

$ \log _{ 10 }{ (x-10) } =1$


$ \log _{ 10 }{ (x-10) } =\log _{ 10 }{ 10 }$

$x-10=10$

$  = 20 $

The value of $7 log _a \displaystyle \frac{16}{15} + 5 log _a \frac{25}{24} + 3 log _a \frac{81}{80}$ is

physics logarithms introduction to logarithm logarithmic notation exponential and logarithms

  1. $log _{a3}$

  2. $log _{a1}$

  3. $log _{a2}$

  4. $log _{a5}$

Show answer
Correct Option: C
Explanation:

We have,

$7log _a\dfrac{16}{15}+5log _a\dfrac{25}{24}+3log _a\dfrac{81}{80}$
$\Rightarrow log _a(\dfrac{16}{15})^7+log _a(\dfrac{25}{24})^5+log _a(\dfrac{81}{80})^3$
$\Rightarrow log _a(\dfrac{16}{15})^7\times (\dfrac{25}{24})^5\times (\dfrac{81}{80})^3$
$\Rightarrow log _a\dfrac{16^3\times 16^4}{5^7\times 3^7}\times \dfrac{5^5\times 5^5}{8^5\times 3^5}\times \dfrac{3^6\times 3^6}{16^3\times 5^3}$
$\Rightarrow log _a\dfrac{16^4}{1\times 1}\times \dfrac{1}{8^5\times 1}\times \dfrac{1}{1}$
$\Rightarrow log _a\dfrac{2^4\times 8^4}{8^5}$
$\Rightarrow log _a\dfrac{16}{8}$
$\Rightarrow log _a2$

Hence, this is the answer.

If $log _{10} x - log _{10} \sqrt x = \displaystyle \frac{2}{log _{10} x}$, then value of x is

physics logarithms introduction to logarithm logarithmic notation exponential and logarithms

  1. $\displaystyle \frac{1}{100}$ or $100$

  2. $\pm$ 2

  3. 10 or $\displaystyle \frac{1}{10}$

  4. 100

Show answer
Correct Option: A
Explanation:

Consider the given equation.

$log _{10}x-log _{10}\sqrt x=\dfrac{2}{log _{10}x}$
$log _{10}\dfrac{x}{\sqrt x}=\dfrac{2}{log _{10}x}$
$log _{10}\sqrt x=\dfrac{2}{log _{10}x}$
$\dfrac{1}{2}log _{10}\ x=\dfrac{2}{log _{10}x}$
$\dfrac{1}{2}(log _{10}\ x)^2=2$
$(log _{10}\ x)^2=4$
$log _{10}\ x=\pm 2$
$x=10^{\pm2}$
$x=100\ or\ \dfrac{1}{100}$

Hence, this is the answer.

If $\displaystyle \frac{log _2 (9 - 2^x)}{3 - x} = 1$, then value of x is

physics logarithms introduction to logarithm logarithmic notation exponential and logarithms

  1. x = 4

  2. x = + 1 or -1

  3. x = $\pm$ 2

  4. x = 0

Show answer
Correct Option: D
Explanation:

We have,

$\dfrac{log _2(9-2^x)}{3-x}=1$
$log _2(9-2^x)=3-x$
$9-2^x=2^{3-x}$                $ .......... (1)$

From option $(D)$
$9-2^0=2^{3-0}$
$9-1=2^3$
$8=8$

Hence, $x=0$ is the root of this equation.

Hence, only option $D$ is correct.

The value of $\log _{ \frac{1}{2} }{ 4 } $ is

physics logarithms introduction to logarithm logarithmic notation exponential and logarithms

  1. $-2$

  2. $0$

  3. $\dfrac{1}{2}$

  4. $2$

Show answer
Correct Option: A
Explanation:

Let $\log _{\frac{1}{2}}4 = x $

Then, $(\dfrac{1}{2})^x = 4 $

$Or, (\dfrac{1}{2})^x =2^2$

$Or, 2^{-x} = 2^2$

$x= -2$

The equation  ${ \left( \log _{ 10 }{ x+2 }  \right)  }^{ 3 }+{ \left( \log _{ 10 }{ x-1 }  \right)  }^{ 3 }={ \left( 2\log _{ 10 }{ x+1 }  \right)  }^{ 3 }$ has

physics logarithms introduction to logarithm logarithmic notation exponential and logarithms

  1. no natural solution

  2. two rational solutions

  3. no prime solution

  4. one irrational solution

Show answer
Correct Option: B,C,D
Explanation:

Let $ \log _{ 10 }{ x+2 } =a$ and $ \log _{ 10 }{ x-1 } =b$
$\therefore a+b=2\log _{ 10 }{ x+1 } $ (from the question)
Thus, the given equation(in the question) reduces to ${a}^{3}+{b}^{3}={(a+b)}^{3}$
$\Rightarrow 3ab(a+b)=0$
$\Rightarrow a=0$ or $b=0$ or $a+b=0$
$\Rightarrow \log _{ 10 }{ x+2 }=0$  or $\log _{ 10 }{ x-1 }=0$ or $2\log _{ 10 }{ x } +1=0$
$\Rightarrow x={10}^{-2}$  or  $x=10$ or  $x={ 10 }^{ -\frac { 1 }{ 2 }  }$
Hence  $x=\left{ \dfrac { 1 }{ 100 },10 ,\dfrac { 1 }{ \sqrt { 10 }  }  \right} $

Which is the correct order for a given number $\alpha$ in increasing order.

maths logarithms more about logarithms a relation between logarithmic functions properties of logarithms

  1. $\log _{2} \alpha, \log _{e} \alpha, \log _{3} \alpha, \log _{10} \alpha$

  2. $\log _{10} \alpha, \log _{3} \alpha, \log _{e} \alpha, \log _{2} \alpha$

  3. $\log _{10} \alpha, \log _{e} \alpha, \log _{2} \alpha, \log _{3} \alpha$

  4. $\log _{3} \alpha, \log _{e} \alpha, \log _{2} \alpha, \log _{10} \alpha$

Show answer
Correct Option: A
Explanation:
We know that, greater the value of base lesser is the value of the logarithm

Also, $10<3<e(2.71)<2$

$\therefore $ For any positive number $\alpha $,

${ log } _{ 2 }\alpha >{ log } _{ e }\alpha >{ log } _{ 3 }\alpha >{ log } _{ 10 }\alpha $

If $\log _3{(\log _3{a})}+\log _{\cfrac{1}{3}}{\left(\log _{\cfrac{1}{3}}{b}\right)}=1$, then the value of $ab^3$ is 

maths logarithms more about logarithms a relation between logarithmic functions properties of logarithms

  1. $9$

  2. $3$

  3. $1$

  4. $\cfrac{1}{3}$

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Correct Option: C