Tag: logarithm and its uses

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Multiple choice logarithm and its uses basic mathematical concepts physics

What is the value of $[\log _{10} (5\log _{10} 100)]^{2}$?

  1. $4$

  2. $3$

  3. $2$

  4. $1$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The value of $[\log _{10}(5\log _{10} 100)]^{2}$ is
$=[\log _{10}(5\log _{10}10^2)]^2$

$=[\log _{10}(10\log _{10}10)]^2$     .....As $\log a^m=m\log a$
$=[\log _{10}(10\times 1)]^2$     ....As $\log _aa=1$
$=[\log _{10}10]^2$
$=[1]^2=1$

Multiple choice logarithm and its uses basic mathematical concepts physics

The value of $x$ which satisfy $log(x+1) = 2logx$ is 

  1. $1$

  2. $\dfrac{\sqrt{5}-1}{2}$

  3. $\dfrac{\sqrt{5}+1}{2}$

  4. $2$
Reveal answer Fill a bubble to check yourself
B,C Correct answer
Explanation

$\log(x+1)=2\log x$


$\Rightarrow \log (x+1)=\log x^2$

$\Rightarrow x^2-x-1=0$

$\Rightarrow x=\cfrac{+1\pm \sqrt{1+4}}{2}=\cfrac{1\pm \sqrt{5}}{2}$

$\Rightarrow x= \left(\cfrac{1+\sqrt{5}}{2}\right),\left(\cfrac{1-\sqrt{5}}{2}\right)$

Multiple choice logarithm and its uses basic mathematical concepts physics

The value of $x$ satisfying the equation $g^{log _3 (log _2 x)} = log _2 x - (log _2 x)^2 + 1$ is 

  1. $0$

  2. $1$

  3. $2$

  4. None
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Here, $g^{\log _{3}(\log _{2}x)}$ is an exponential function and $\log _{2}{x}-\left(\log _{2}{x}\right)^{2}+1$ is a quadratic with imaginary roots.

The two can be equal when both side become $0,1$. Since, right hand side can become zero at imaginary point. We, only consider, then the two side become $1$.
$\log _{2}{x}-\left(\log _{2}{x}\right)^{2}+1=1$
$\Rightarrow \left(\log _{2}{x}\right)^{2}-\left(\log _{2}{x}\right)=0$
$\Rightarrow \left(\log _{2}{x}\right)\left(\log _{2}{x}-1\right)=0$
$\Rightarrow \log _{2}{x}=0$ and $\log _{2}{x}=1$
$\Rightarrow x=1,2$
But $x\neq 1$ as in $g^{\log _{3}(\log _{2}x)}$ it become invalid hence, $x=2$ satisfy the relation.

Multiple choice logarithm and its uses basic mathematical concepts physics

If $2y = log(12-5x-3x^2)$ takes all real values then $x$ belongs to 

  1. $(-3, 5/3)$

  2. $(-3, 3)$

  3. $(-3, 4/3)$

  4. None
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$2y=\log\left(12-5x-3x^2\right)$

$(12-5x-3x^2)>0$
$3x^2+5x-12<0$
$3x^2+9x-4x-12<0$
$3x\left(x+3\right)-4\left(x+3\right)<0$
$\left(x+3\right)\left(3x-4\right)<0$
$x\epsilon \left(-3,{4/3}\right)$

Multiple choice logarithm and its uses basic mathematical concepts physics

Evaluate the expression by using logarithm tables: $ \dfrac{(17.42)^{2/{3}}\times 18.42}{\sqrt{126.37}}$

  1. $11.01$

  2. $12.01$

  3. $13.01$

  4. $14.01$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let $x= \dfrac{(17.42)^{2/{3}}\times 18.42}{\sqrt{126.37}}$
Taking logarithm on both sides,
$ \log { x } =\log { (17.42)^{ 2/{ 3 } } } +\log { 18.42 } -\log { \sqrt { 126.37 }  } $

$\log { x } =\dfrac { 2 }{ 3 } \log { 17.42 } +\log { 18.42 } -\dfrac { 1 }{ 2 } \log { (126.37) } $

$ \log { x } =\dfrac { 2 }{ 3 } \log { (1.742\times 10) } +\log { (1.842\times 10) } -\dfrac { 1 }{ 2 } \log { (1.264\times { 10 }^{ 2 }) } $

$ \log { x } =\dfrac { 2 }{ 3 } { (1.2410) } + { 1.2653 } -\dfrac { 1 }{ 2 } { (2.1018) } $

$\log { x } =0.8273+1.2653-1.0509$

$\log { x } =1.0417$
$\Rightarrow x= \text{antilog }(1.0417)$
$\Rightarrow x = 11.01$

Multiple choice logarithm and its uses basic mathematical concepts physics

Let $a = \log 3\log _32$. An integer k satisfying  $1< 2^{(-k+3^{-a})} < 2,$  must be less than ____.

  1. $1.25766$

  2. $2.256$

  3. $3$

  4. $1$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given a = log3(log3(2)). The inequality 1 < 2^(-k + 3^-a) < 2 simplifies to 0 < -k + 3^-a < 1. Since 3^-a = 3^-log3(log3(2)) = 1/log3(2) = log2(3) approx 1.585. So 0 < -k + 1.585 < 1, which means 0.585 < k < 1.585. Thus k must be less than 1.585.