Tag: logarithms

Quizzes (3) Questions (108)

Questions Related to logarithms

Multiple choice physics logarithms introduction to logarithm logarithmic notation 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. $\displaystyle \log _5 3 = 3$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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

Multiple choice physics logarithms introduction to logarithm logarithmic notation exponential and logarithms

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. $\displaystyle \log _{78} 12 = \frac {1}{3}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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

Multiple choice physics logarithms introduction to logarithm logarithmic notation exponential and logarithms

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
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$\displaystyle \log F = \log  \: G + \log  \: m _1 + \log  \: m _2 - 2 \: \log  \: d$

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

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

$\therefore F=\cfrac{Gm _1m _2}{d^2}$
Multiple choice physics logarithms introduction to logarithm logarithmic notation exponential and logarithms

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
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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