Tag: exponential and logarithms

Questions Related to exponential and logarithms

The logarithm form of $\displaystyle 5^3 = 125$ is equal to

  1. $\displaystyle \log _5 125 = 3$

  2. $\displaystyle \log _5 125 = 5$

  3. $\displaystyle \log _3 125 = 5$

  4. $\displaystyle \log _5 3 = 3$


Correct Option: A
Explanation:

$5^3=125$
Taking log on both sides, we get
$3\log 5 = \log 125$
$\log _5 125 = 3$         ...(since $\dfrac{\log a}{\log b} = \log _ba$)

The logarithmic form of $\displaystyle (81)^{\frac {3}{4}} = 27$ is

  1. $\displaystyle \log _{66} 36 = \frac {2}{9}$

  2. $\displaystyle \log _{81} 27 = \frac {3}{4}$

  3. $\displaystyle \log _{16} 33 = \frac {7}{2}$

  4. $\displaystyle \log _{78} 12 = \frac {1}{3}$


Correct Option: B
Explanation:

$81^{\frac{3}{4}}=27$
Taking log on both sides
$\dfrac{3}{4}log81=log27$
$log _{81}27= \dfrac{3}{4}$.....(since $\dfrac{loga}{logb}=log _ba$)

Given $\displaystyle 3^{x} = \frac {1}{9}$ then $x=?$

  1. $-1$

  2. $-2$

  3. $1$

  4. $2$


Correct Option: B
Explanation:

$3^{-2}=\dfrac{1}{9}$
Taking log on both sides
$-2log3= log\dfrac{1}{9}$
$log _3\dfrac{1}{9}= -2$.....(since $\dfrac{loga}{logb}=log _ba$)

Express in logarithmic form and find x: $\displaystyle 10^{x} = 0.001$ (i.e base 10)

  1. $3$

  2. $-3$

  3. $2$

  4. $-2$


Correct Option: B
Explanation:

$10^{-3}=0.001$
Taking log on both sides
$-3log10= log0.001$
$log _{10}0.001= -3$.....(since $\dfrac{loga}{logb}=log _ba$)

State whether true or false:
$\displaystyle \log F = \log : G + \log : m _1 + \log : m _2 - 2 \log : d$ gives $\displaystyle F = G \frac {m _2m _1}{d^2}$.

  1. True

  2. False


Correct Option: A
Explanation:
$\displaystyle \log F = \log  \: G + \log  \: m _1 + \log  \: m _2 - 2 \: \log  \: d$

$\therefore \log F= \log G+\log m _1+\log m _2-\log d^2$....($\log a^b=b\log a$)

$\therefore \log F= \log \cfrac{Gm _1m _2}{d^2}$.....($\log a.b=\log a+\log b, \log \cfrac{a}{b}=\log a-\log b$)

$\therefore F=\cfrac{Gm _1m _2}{d^2}$

State whether true or false:
$\displaystyle \log : V = 2 \log : 2 - \log : 3 + \log : \pi + 3 \log : r$ gives $\displaystyle V = \frac {4}{3} \pi r^3$

  1. True

  2. False


Correct Option: A
Explanation:

$\displaystyle \log : V = 2 : \log : 2 - : \log : 3 + : \log : \pi + 3 : \log : r$
$\therefore \log V = \log 2^2-\log 3+\log \pi+ \log r$....($\log a^b=b \log a$)
$\therefore \log V= \log \cfrac{4}{3}\pi r^3$.....($\log a.b = \log a + \log b, \log \cfrac {a}{b} = \log a - \log b$)
$\therefore V=\cfrac{4}{3}\pi r^3$

The logarithm form of $10^{-3} = 0.001$ is $\log _{10} 0.001 = -m$, then value of $m$ is 

  1. $-1$

  2. $-2$

  3. $3$

  4. $-4$


Correct Option: C
Explanation:
The logarithm of the $10^{-3}=\frac{1}{1000}=0.0001$
So $log _{10}0.0001=-3$
$log _{10}0.0001=-m$
Then m=3

The value of $\displaystyle \log _{10}0.001 $ is equal to

  1. $-3$

  2. $3$

  3. $-2$

  4. $2$


Correct Option: A
Explanation:

$ \ \log _{ 10 }{ 0.001 } =x\ 0.001 = { 10 }^{ x }\ \dfrac { 1 }{ 1000 } = { 10 }^{ -3 }\ x=-3  $

The value of $\log _{0.5}16$ is equal to

  1. $-4$

  2. $-1$

  3. $-2$

  4. $0$


Correct Option: A
Explanation:

$ \log _{ 0.5 }{ 16 } = x $


$ 16 = {0.5 }^{ x }$

$ { 2 }^{ 4 } = { 2 }^{ -1x }$

$ x = -4\ \ 
$

The logarithm of $0.001$ to the base $10$ is equal to

  1. $5$

  2. $-1$

  3. $6$

  4. $-3$


Correct Option: D
Explanation:

$0.001$ to the base $10$
So, as per 
logarithm,
$\log _{ 10 }{ 0.001 } =\log _{ 10 }{ { 10 }^{ -3 } } $
$\log _{ 10 }{ 0.001 } =-3\log _{ 10 }{ { 10 } } $
$\log _{ 10 }{ 0.001 } =-3\times 1$
$\log _{ 10 }{ 0.001 } =-3$