Tag: exponential and logarithms

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Questions Related to exponential and logarithms

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

$\log V = 2 \log 2 - \log 3 + \log \pi + 3 \log r$ can be expressed as

  1. $V = \dfrac{4}{3} \pi r^{3}$
  2. $ V = \dfrac{2}{3} \pi r^{3}$
  3. $ V = \dfrac{4}{3} \pi r$
  4. $ V = \dfrac{2}{3} \pi r$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$ \log { V }  = 2\log { 2 } -\log { 3 } +\log { \pi  }  + 3\log { r }$
$ \log V = \log ({ 2 }^{ 2 }\times \pi \times { r }^{ 3 }) - \log { 3 }$
$ \log V = \log \dfrac { 4\pi { r }^{ 3 } }{ 3 } $
Removing log
$V = \dfrac {4}{3} \pi r^3$
Multiple choice physics logarithms introduction to logarithm logarithmic notation exponential and logarithms

Which of the following is true for $\log _25$?

  1. An integer

  2. A rational number

  3. An irrational number

  4. A whole number
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let us assume that $\log _{ 2 }{ 5 } =\frac { p }{ q } $ , where $p,q$ are integers

We have $5=2^{\frac{p}{q}}$
$\Rightarrow 5^{q}=2^{p}$
This suggest that $5$ and $2$ are not mutually prime , But $2$ and $5$ are mutually prime
Therefore $\log _{ 2 }{ 5 } $ is an irrational number

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

The value of $7 log _a \displaystyle \frac{16}{15} + 5 log _a \frac{25}{24} + 3 log _a \frac{81}{80}$ is

  1. $log _{a3}$
  2. $log _{a1}$
  3. $log _{a2}$
  4. $log _{a5}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

We have,

$7log _a\dfrac{16}{15}+5log _a\dfrac{25}{24}+3log _a\dfrac{81}{80}$
$\Rightarrow log _a(\dfrac{16}{15})^7+log _a(\dfrac{25}{24})^5+log _a(\dfrac{81}{80})^3$
$\Rightarrow log _a(\dfrac{16}{15})^7\times (\dfrac{25}{24})^5\times (\dfrac{81}{80})^3$
$\Rightarrow log _a\dfrac{16^3\times 16^4}{5^7\times 3^7}\times \dfrac{5^5\times 5^5}{8^5\times 3^5}\times \dfrac{3^6\times 3^6}{16^3\times 5^3}$
$\Rightarrow log _a\dfrac{16^4}{1\times 1}\times \dfrac{1}{8^5\times 1}\times \dfrac{1}{1}$
$\Rightarrow log _a\dfrac{2^4\times 8^4}{8^5}$
$\Rightarrow log _a\dfrac{16}{8}$
$\Rightarrow log _a2$

Hence, this is the answer.

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

If $log _{10} x - log _{10} \sqrt x = \displaystyle \frac{2}{log _{10} x}$, then value of x is

  1. $\displaystyle \frac{1}{100}$ or $100$
  2. $\pm$ 2
  3. 10 or $\displaystyle \frac{1}{10}$

  4. 100
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Consider the given equation.

$log _{10}x-log _{10}\sqrt x=\dfrac{2}{log _{10}x}$
$log _{10}\dfrac{x}{\sqrt x}=\dfrac{2}{log _{10}x}$
$log _{10}\sqrt x=\dfrac{2}{log _{10}x}$
$\dfrac{1}{2}log _{10}\ x=\dfrac{2}{log _{10}x}$
$\dfrac{1}{2}(log _{10}\ x)^2=2$
$(log _{10}\ x)^2=4$
$log _{10}\ x=\pm 2$
$x=10^{\pm2}$
$x=100\ or\ \dfrac{1}{100}$

Hence, this is the answer.

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

If $\displaystyle \frac{log _2 (9 - 2^x)}{3 - x} = 1$, then value of x is

  1. x = 4

  2. x = + 1 or -1
  3. x = $\pm$ 2

  4. x = 0
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

We have,

$\dfrac{log _2(9-2^x)}{3-x}=1$
$log _2(9-2^x)=3-x$
$9-2^x=2^{3-x}$                $ .......... (1)$

From option $(D)$
$9-2^0=2^{3-0}$
$9-1=2^3$
$8=8$

Hence, $x=0$ is the root of this equation.

Hence, only option $D$ is correct.

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

The equation  ${ \left( \log _{ 10 }{ x+2 }  \right)  }^{ 3 }+{ \left( \log _{ 10 }{ x-1 }  \right)  }^{ 3 }={ \left( 2\log _{ 10 }{ x+1 }  \right)  }^{ 3 }$ has

  1. no natural solution

  2. two rational solutions

  3. no prime solution

  4. one irrational solution
Reveal answer Fill a bubble to check yourself
B,C,D Correct answer
Explanation

Let $ \log _{ 10 }{ x+2 } =a$ and $ \log _{ 10 }{ x-1 } =b$
$\therefore a+b=2\log _{ 10 }{ x+1 } $ (from the question)
Thus, the given equation(in the question) reduces to ${a}^{3}+{b}^{3}={(a+b)}^{3}$
$\Rightarrow 3ab(a+b)=0$
$\Rightarrow a=0$ or $b=0$ or $a+b=0$
$\Rightarrow \log _{ 10 }{ x+2 }=0$  or $\log _{ 10 }{ x-1 }=0$ or $2\log _{ 10 }{ x } +1=0$
$\Rightarrow x={10}^{-2}$  or  $x=10$ or  $x={ 10 }^{ -\frac { 1 }{ 2 }  }$
Hence  $x=\left{ \dfrac { 1 }{ 100 },10 ,\dfrac { 1 }{ \sqrt { 10 }  }  \right} $