Tag: introduction to logarithm

Questions Related to introduction to logarithm

If mantissa of logarithm of 719.3 to the base 10 is 0.8569 , then mantissa of logarithm  of 71.93 is

physics logarithms introduction to logarithm logarithmic notation exponential and logarithms

  1. 0.8569

  2. $\overline 1 .8569$

  3. 1.8569

  4. 0.1431

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Correct Option: A
Explanation:

Mantissa of logarithm of $719.3$ to base $10$ is $0.8569$

Then, mantissa of logarithm of $71.93$ is also $0.8569$
As, $\log _{10}{(719.3)}=2+(0.8569)$(mantissa)
So, $\log _{10}{(71.93)}=1+(0.8596)$ (mantissa)

If $2\log y -\log x -3=0$, express $x$ in terms of $y.$

physics logarithms introduction to logarithm logarithmic notation exponential and logarithms

  1. $x=\dfrac{y^2}{e^3}$

  2. $x=\dfrac{y^2}{e^2}$

  3. $x^2=\dfrac{y^2}{e^3}$

  4. $x=\dfrac{y^3}{e^3}$

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Correct Option: A
Explanation:
$\Rightarrow$$\log { { y }^{ 2 } } -\log { x } -\log { { e }^{ 3 } } =0$.......$\log e=1$

$\Rightarrow$$ \log { x } =\log { \left (\cfrac { { y }^{ 2 } }{ { e }^{ 3 } } \right ) } $

$\Rightarrow$$ x=\cfrac { { y }^{ 2 } }{ { e }^{ 3 } } $

If $2\log y -\log x-3=0$ express $x$ in terms of $y.$

physics logarithms introduction to logarithm logarithmic notation exponential and logarithms

  1. $x^2=1000y$

  2. $x^2= \dfrac{y^2}{e^3}$

  3. $y^2= \dfrac{x}{1000}$

  4. $y^2= 1000x$

Show answer
Correct Option: B
Explanation:
Given: $2\log y -\log x-3=0$

$\log { { y }^{ 2 } } -\log { x } -3\log { { e }=0 } $.......$(\log e=1)$

$\log { { y }^{ 2 } } -\log { x } -\log { { e }^{ 3 }=0 } $

$ \log { x } =\log { { y }^{ 2 } } -\log { { e }^{ 3 } } =\log { \left (\cfrac { { y }^{ 2 } }{ { e }^{ 3 } } \right ) } $

$ x=\cfrac { { y }^{ 2 } }{ { e }^{ 3 } } $

If $2x^{{log _4}^3}+3^{\log _4x}=27$, then x is equal to?

physics logarithms introduction to logarithm logarithmic notation exponential and logarithms

  1. $2$

  2. $4$

  3. $8$

  4. $16$

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Correct Option: D
Explanation:
$2\times { x }^{ { log } _{ 4 }3 }+{ 3 }^{ { log } _{ 4 }x }=27$

${ 2\times3 }^{ { log } _{ 4 }x }+{ 3 }^{ { log } _{ 4 }x }=27$

Let ${ 3 }^{ { log } _{ 4 }x }=t$
$2t+t=27$
$3t=27$
$t=9$

${ 3 }^{ { log } _{ 4 }^{ x } }={ 3 }^{ 2 }=9$

So,  ${ log } _{ 4 }x=2$

So,  $x={ 2 }^{ 4 }=16$

$\log _{ 4 }{ 18 } $ is

physics logarithms introduction to logarithm logarithmic notation exponential and logarithms

  1. A rational number

  2. An irrational number

  3. A prime number

  4. None of these

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Correct Option: B
Explanation:
$log _418=\dfrac{log18}{log4}$
$=\dfrac{log2+log9}{2log2}$
$=\dfrac{log2+2log3}{2log2}$
$=\dfrac{log2}{2log2}+\dfrac{2log3}{2log2}$
$=\dfrac{1}{2}+log _23$
Since $log _23$ is irrational
Hence $=\dfrac{1}{2}+log _23$ is irrational.
Therefore
$log _418$ is irrational number

The value of x, for which the 6th term in the expansion of $\left{ { 2 }^{ { log } _{ 2 }\sqrt { \left( { 9 }^{ x-1 }+7 \right)  }  }+\dfrac { 1 }{ { 2 }^{ { \left( 1/5 \right) log } _{ 2 }\left( { 3 }^{ x-1 }+1 \right)  } }  \right} ^{ 7 }$ is 84, is equal to 

physics logarithms introduction to logarithm logarithmic notation exponential and logarithms

  1. 4

  2. 3

  3. 2

  4. 1

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Correct Option: C

If x = ${ log } _{ 3 }243,y={ log } _{ 2 }64,$, Then $\sqrt { x-2\sqrt { y }  } $ is 

physics logarithms introduction to logarithm logarithmic notation exponential and logarithms

  1. $\sqrt { 5-2\sqrt6 }$

  2. $2-\sqrt { 3 } $

  3. $\sqrt { 3 } -\sqrt { 2 } $

  4. $\sqrt { 3 } -4$

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Correct Option: A
Explanation:

We have,

$ x={{\log } _{3}}243\,\,......\,\,\left( 1 \right) $

$ y={{\log } _{2}}64\,\,......\,\,\left( 2 \right) $

Applying rule,

${{\log } _{b}}x=y\,\Rightarrow {{b}^{y}}=x$

So,

From equation (1) and (2) to,

$ x={{\log } _{3}}243 $

$ \Rightarrow {{3}^{x}}=243 $

$ \Rightarrow {{3}^{x}}={{3}^{5}} $

$ \Rightarrow x=5 $

Now,

$ {{\log } _{2}}64=y $

$ \Rightarrow {{2}^{y}}=64 $

$ \Rightarrow {{2}^{y}}={{2}^{6}} $

$ \Rightarrow y=6 $

Now,

$ \sqrt{x-2\sqrt{y}} $

$ =\sqrt{5-2\sqrt{6}} $

Hence, this is the answer.

Logarithmic form of  $3 \sqrt { 8 } = 2$  is

physics logarithms introduction to logarithm logarithmic notation exponential and logarithms

  1. $\log _ { 8 } 2 = \dfrac { 1 } { 3 }$

  2. $\log _ { 2 } 8 = \dfrac { 1 } { 3 }$

  3. $\log _ { \frac { 1 } { 3 } } 8 = 2$

  4. $\log _ { \frac { 1 } { 3 } } 2 = 0$

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Correct Option: A

Number of solutions of $\log _{4}{\left(x-1\right)}=\log _{2}{\left(x-3\right)}$

physics logarithms introduction to logarithm logarithmic notation exponential and logarithms

  1. $1$

  2. $2$

  3. $3$

  4. $4$

Show answer
Correct Option: B
Explanation:
$\log _{4}{\left(x-1\right)}=\log _{2}{\left(x-3\right)}$
$\Rightarrow\,\log _{{2}^{2}}{\left(x-1\right)}=\log _{2}{\left(x-3\right)}$
$\Rightarrow\,\dfrac{1}{2}\log _{2}{\left(x-1\right)}=\log _{2}{\left(x-3\right)}$
$\Rightarrow\,\log _{2}{\left(x-1\right)}=2\log _{2}{\left(x-3\right)}$
$\Rightarrow\,\left(x-1\right)={\left(x-3\right)}^{2}$
$\Rightarrow \,x-1={x}^{2}-6x+9$
$\Rightarrow \,{x}^{2}-6x-x+9+1=0$
$\Rightarrow \,{x}^{2}-7x+10=0$
$\Rightarrow \,{x}^{2}-2x-5x+10=0$
$\Rightarrow \,x\left(x-2\right)-5\left(x-2\right)=0$
$\Rightarrow \,\left(x-2\right)\left(x-5\right)=0$
$\therefore\,x=2,\,5$
Number of solutions$=2$

The value of x, which satisfies the equation $2 \log _ { 2 } \left( \log _ { 2 } x \right) + \log _ { 12 } \left( \log _ { 2 } ( 2 \sqrt { 2 } x ) \right) = 1$ is greater

physics logarithms introduction to logarithm logarithmic notation exponential and logarithms

  1. 10

  2. 11

  3. 7

  4. 9

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Correct Option: A