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Questions Related to physics

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.

Multiple choice logarithm and its uses basic mathematical concepts physics

If $a=\log _35 $ and $b= \log _725$ then correct option is:

  1. $a &lt; b$

  2. $ a &gt; b$

  3. $a= b$

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$a=\log _{ 3 }{ 5 } =\cfrac { \log { 5 }  }{ \log { 3 }  } ,b=\cfrac { \log { 25 }  }{ \log { 7 }  } =\cfrac { \log { 5^2 }  }{ \log { 7 }  }=\cfrac { 2\log { 5 }  }{ \log { 7 }  } $


$ \cfrac { a }{ b } =\cfrac { \log { 5 }  }{ \log { 3 }  } \times \cfrac { \log { 7 }  }{ 2\log { 5 }  } =\cfrac { 1 }{ 2 } \log _{ 3 }{ 7 } =\log _{ 3 }{ (\sqrt { 7 } ) } $


$ Now,\sqrt { 7 } <3,so\quad \cfrac { a }{ b } <1$

$ \cfrac { a }{ b } <1\  =>a<b$

Multiple choice logarithm and its uses basic mathematical concepts physics

The value of ${ \left( 0.05 \right)  }^{ \log _{ \sqrt { 20 }  }{ \left( 0.1+0.01+0.001+.... \right)  }  }$ is 

  1. $81$

  2. $\cfrac{1}{81}$

  3. $20$

  4. $\cfrac{1}{20}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$0.1+0.01+0.0001+..... \Rightarrow G.P$
$=(0.1)(1-0.1)\quad S _{\infty}=-\dfrac {a}{1-r}$
$=\dfrac {0.1}{0.9}=\dfrac {1}{9}$
$\therefore \ (0.05)\log _\sqrt {20} (0.1+0.01+....)$
$=\left (\dfrac {1}{20}\right)\log \sqrt {20}^{1/9}$
$=\left (\dfrac {1}{9}\right) \log \sqrt {20}^{1/20}$
$=\left (\dfrac {1}{20}\right) \log \sqrt {20}^{(\sqrt {20})^{-2}}=\left (\dfrac {1}{20}\right)^{-2}$
$=81$
Multiple choice logarithm and its uses basic mathematical concepts physics

The equation ${ x }^{ \cfrac { 3 }{ 4 } { \left( \log _{ x }{ x }  \right)  }^{ 2 }+\log _{ x }{ x } -\cfrac { 5 }{ 4 }  }=\sqrt { 2 } $ has

  1. at least one real solution

  2. exactly three solutions

  3. exactly one irrational solution

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

The equation simplifies significantly because log_x(x) = 1. The equation becomes x^(3/4 * 1^2) + 1 - 5/4 = sqrt(2), which is x^(3/4) - 1/4 = sqrt(2). This has a real solution for x.