Tag: logarithm and its uses

Questions Related to logarithm and its uses

The antilog of the number $0.2015$ is equal to

logarithm and its uses basic mathematical concepts physics

  1. $\overline {1}.1591$

  2. $1.591$

  3. $0.01591$

  4. $15.91$

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Correct Option: B
Explanation:

The number before decimal point is $0,$ so decimal point will be after $1$ digits.

Value of $0.2015$ from antilog table $=1589+2+=1591$
Now place a decimal point after $1$ digits of the number from left we get, $1.591$
Antilog $0.2015=1.591$
Hence, B is the correct option.

If there are $n$ zeros after the decimal point, then the characteristic of that number will be

logarithm and its uses basic mathematical concepts physics

  1. $n+1$

  2. $-n+1$

  3. $-(n+1)$

  4. $n-1$

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Correct Option: C
Explanation:

For number less than $1,$ if there are $n$ zeroes after the decimal point, then the characteristics of that number will be $-(n+1).$

and For number  greater than $1,$ if  there are $n$ number of zeroes are on the left sides of the digits then characteristics will be $(n+1).$
Hence, C is the correct option.

Evaluate using log tables: $\sqrt [3] {\dfrac {16.23}{426.8}}$

logarithm and its uses basic mathematical concepts physics

  1. $0.6332$

  2. $0.3632$

  3. $0.3362$

  4. $0.3624$

Show answer
Correct Option: C
Explanation:

$\log { y } =\cfrac { 1 }{ 3 } \left[ \log { (16.23) } -\log { (426.8) }  \right] \ \log { y } =\cfrac { 1 }{ 3 } \left[ 1.2103-2.63022 \right] \ \log { y } =-0.4733\ y={ 10 }^{ -0.4733 }=0.3362$

If $f(x) = \log x$, then $f^{-1}x  $ is

logarithm and its uses basic mathematical concepts physics

  1. $\log\dfrac1x$

  2. $\log x^2$

  3. $anti\log(x)$

  4. $anti\log\dfrac1x$

Show answer
Correct Option: C
Explanation:

Let,$f(x)=y=\log x$

$\Rightarrow f(x)=y=\log x$
$\Rightarrow $ anti$\log (y)=x$
Therefore, Inverse of $f(x)$ i.e. $f^{-1}(x)=$anti$\log (x)$
Hence, C is the correct option.

The antilog of $(4.8779)$ will be 

logarithm and its uses basic mathematical concepts physics

  1. $7500$

  2. $750$

  3. $75500$

  4. $750000$

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Correct Option: C
Explanation:

The number before decimal point is $4,$ so decimal point will be after $5$ digits.

Value of $0.8779$ from antilog table $=7534+16=7550$
Now place a decimal point after $5$ digits of the number from left we get, $75500$
Antilog $4.8779=75500$

The value of $\log _{10} 8$ is equal to

logarithm and its uses basic mathematical concepts physics

  1. $.903$

  2. $3.901$

  3. $.301$

  4. None of the above

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Correct Option: A
Explanation:

$\log _{10}8=\log _{10}2^3=3\log _{10}2$

$=3\times 0.301$   $(\log _{10}2=0.301)$
$=0.903$
Hence, D is the correct option.

Find the value of $\dfrac {\log _{10} 72}{\log _{10} 8}$ using log table

logarithm and its uses basic mathematical concepts physics

  1. $\log _{10} 9$

  2. $1+\dfrac{.954}{.903}$

  3. $2$

  4. $\dfrac{.903+.954}{.954}$

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Correct Option: B
Explanation:
$\dfrac{\log _{10}{72}}{\log _{10}{8}}=\dfrac{\log _{10}{(8\times9)}}{\log _{10}{8}}=\dfrac{\log _{10}{8}+\log _{10}{9}}{\log _{10}{8}}$
$=1+\dfrac{\log _{10}{3^{2}}}{\log _{10}{2^{3}}}=1+\dfrac{2\log _{10}{3}}{3\log _{10}{2}}$
$=1+\dfrac{2.(0.477)}{3.(0.301)}=1+\dfrac{(0.954)}{(0.903)}$

Find the value of $\log _{10} 72$ using log table

logarithm and its uses basic mathematical concepts physics

  1. $0.901+0.909$

  2. $0.903+0.954$

  3. $1.890$

  4. $2.104$

Show answer
Correct Option: B
Explanation:
$\log _{10}{72}=\log _{10}(2^{3}.3^{2})$
$=\: \log _{10}{2^{3}}+\log _{10}{3^{2}}$
$=\: 3.\log _{10}{2}+2.\log _{10}{3}$
$=\: 3(0.301)+2(0.4771)$
$=\: 0.903+0.954$

Find $AntiLog(.2817)$.

logarithm and its uses basic mathematical concepts physics

  1. $1.70$

  2. $2.19$

  3. $1.91$

  4. $1.99$

Show answer
Correct Option: C
Explanation:

We know that $AntiLog(x)=10^{x}$

$\Rightarrow AntiLog(0.2817)=10^{0.2817}=1.91$
This exact value should be find by using anti logarithm table.
Therefore the correct option is $C$

Find the value of $\log _{10} {\dfrac{64^{2.1}\times 81^{4.2}}{49^{3.4}}}$ using log table

logarithm and its uses basic mathematical concepts physics

  1. $2.1 \times 6 \times .303+ 4.2 \times 2 \times .854- 3.4 \times 2 \times .745$

  2. $2.1 \times .303+ 4.2 \times .954- 3.4 \times .845$

  3. $2.1 \times 6 \times .303- 4.2 \times 2 \times .954+3.4 \times 2 \times .845$

  4. $2.1 \times 6 \times .303+ 4.2 \times 2 \times .954- 3.4 \times 2 \times .845$

Show answer
Correct Option: D
Explanation:
$\log _{10}{64^{2.1}}+\log _{10}{81^{4.2}}-\log _{10}{49^{3.4}}$
$= 2.1\log _{10}{64}+4.2\log _{10}{81}-3.4\log _{10}{49}$
$=2.1\log _{10}{2^{6}}+4.2\log _{10}{3^{4}}-3.4\log _{10}{7^{2}}$
$=2.1\times6\times(0.303)+4.2\times4\times(0.477)-3.4\times2\times(0.845)$
$=2.1\times6\times(0.303)+4.2\times2\times(0.954)-3.4\times2\times(0.845)$