Tag: basic mathematical concepts

Questions Related to basic mathematical concepts

Let $x = (0.15)^{20}$. Find the characteristic in the logarithm of $x$ to the base $10$.

logarithm and its uses basic mathematical concepts physics

  1. $17$

  2. $21$

  3. $-21$

  4. $-17$

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Correct Option: D
Explanation:

Given $x=(0.15)^{20}$

By applying $\log$ on both sides , we get $\log _{ 10 }{ x } =20\log _{ 10 }{ (0.15) } =-16.478$
$\Rightarrow \log _{ 10 }{ x } =-17+0.5218$
The integral part of $\log _{ 10 }{ x } $ is called characteristic
Therefore the characteristic of given number is $-17$
So option $D$ is correct

Find the value of ${\log _{10} 72} + {\log _{10} {\dfrac{1}{8}}}$ using log table

logarithm and its uses basic mathematical concepts physics

  1. $0.903$

  2. $0.303$

  3. $0.954$

  4. $1.234$

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Correct Option: C
Explanation:
$\log _{10}{72}+\log _{10}\left (\dfrac{1}{8}  \right )$
$=\: \log _{10}{\left (72\times \dfrac{1}{8}  \right )}=\log _{10}{9}=\log _{10}{3^{2}}$
$=\: 2.\log _{10}{3}=2(0.477)=0.954$

If $x^2+y^2=25$ , then $log _5 \begin {bmatrix} Max (3x+4y) \end {bmatrix}$ is

logarithm and its uses basic mathematical concepts physics

  1. $2$

  2. $3$

  3. $4$

  4. $5$

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Correct Option: A
Explanation:
$log _5(3x+4y)$

Let $s=3x+4y$
given $x^2+y^2=25$
then $S=3x+4 \sqrt{25-x^2}$
$\dfrac{ds}{dx}=3+4 \dfrac{1}{2\sqrt(25-x^2)}$
$3=\dfrac{4x}{\sqrt{25-x^2}}$
$9(25-x^2)=16x^2$
$x=\pm 3$

and $x^2+y^2=25$
$y^2=25-x^2$
$y=\pm 4$
$\dfrac{d^2s}{dx^2}<0$ ; At $x=3 \,and\, y=4$

$S=3x+4y=3(3)+4(4)=25$
$log _5 (3x+4y)=log _5(s)=log _5(25)=log _5(5^2)=2$

If the mantissa of $\log 2125 =3.3275$, find the mantissa of $\log21.25$

logarithm and its uses basic mathematical concepts physics

  1. $1.3273$

  2. $2.3273$

  3. $0.3273$

  4. $32.2321$

Show answer
Correct Option: A
Explanation:

Given that:

$\log2125=3.3273$
Now, $\log21.25=\log\cfrac{2125}{100}$
$=\log 2125-\log100$
$=3.3273-\log10^2$
$=3.3273-2$
$=1.3273$

The logarithm of $0.0625$ to the base $2$ is:

logarithm and its uses basic mathematical concepts physics

  1. $0.025$

  2. $0.25$

  3. $5$

  4. $-4$

  5. $-2$

Show answer
Correct Option: D
Explanation:

$\log _{2}{\cfrac{625}{10000}}$

$=\log _{2}{\cfrac{1}{16}}$
$=\log _{2}{(16)^{-1}}$
$=\log _{2}{(2)^{-4}}$
$=-4\log _{2}{2}$
$= -4$

The number of zeros between the decimal point and first significant digit of ${\left(0.036\right)}^{16}$ where $log2=0.301$ and $log3=0.477$

logarithm and its uses basic mathematical concepts physics

  1. $21$

  2. $22$

  3. $23$

  4. $24$

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Correct Option: C
Explanation:

$y=(0.036)^{16}$

$\Rightarrow \log (y)= 16 log(0.036)$
$\Rightarrow \log (y)= 16 \log(\cfrac{36}{1000})=16[\log 36- 3\log _{10}{10}]$
$\Rightarrow \log(y)= 16[\log (2^2\times 3^2)-3]=16[2(\log 2+\log 3)-3]$
$\Rightarrow \log(y)=16[2(0.301+0.477)-3]=-23.104<-23$
So, that number has $23$ zeros after this point.

Given $log _{10}2=a$ and $log _{10}3=b$, if $3x+2=25$, the value of x in terms of $a$ and $b$ is $x=(10^{k}+1)$. K=?

logarithm and its uses basic mathematical concepts physics

  1. $\dfrac{a-1}{b}$

  2. $a-b+1$

  3. $\dfrac{1+a}{b}$

  4. $\dfrac{b}{1-a}$

Show answer
Correct Option: B
Explanation:

$\log _{10}{2}=a$

$\Rightarrow 2=(10^a)$
$\log _{10}{3}=b$
$\Rightarrow 3=(10^b)$
$3x+2=25$
$x=\cfrac{23}{3}=(2\times \cfrac{10}{3}+1)=10\cfrac{(10)^a}{(10)^b}+1$
$x=10^{(a+1-b)}+1$
$x=(10^{a-b+1}+1)$

Let $N=\dfrac{\log _{3}135}{\log _{15}3}-\dfrac{\log _{3}5}{\log{405}3}$, then $N$ is

logarithm and its uses basic mathematical concepts physics

  1. a natural number

  2. a prime number

  3. a rational number

  4. an integer

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Correct Option: A,B,C,D

If $x=198!$ then value of the expression $\dfrac {1}{\log _{2}x}+\dfrac {3}{\log _{2}x}+...\dfrac {198}{\log _{2}x}$ equals ?

logarithm and its uses basic mathematical concepts physics

  1. $-1$

  2. $0$

  3. $1$

  4. $198$

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Correct Option: C

The value of $\dfrac{log _2 24}{log _{96} 2}-\dfrac{log _2192}{log _{12}{2}}$ is

logarithm and its uses basic mathematical concepts physics

  1. $3$

  2. $0$

  3. $2$

  4. $1$

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Correct Option: A
Explanation:
Consider
$\dfrac{log _2 24}{log _{96} 2}-\dfrac{log _2192}{log _{12}{2}}\\$
$=\dfrac{log24.log96-log192log12}{(log2)^2}$
$=\dfrac{log(2^3 \times 3)log(2^5\times 3)-log(2^6\times3)log(2^2\times3)}{(log2)^2}$
$=\dfrac{(3log2+log3)(5log2+log3)-(6log2+log3)(2log2+log3)}{(log2)^2}$
$=\dfrac{15(log2)^2-12(log2)^2}{(log2)^2}$
$=3\dfrac{(log2)}{log2}$
$=3$
Option A is the correct answer.