Tag: basic mathematical concepts

Quizzes (8) Questions (78)

Questions Related to basic mathematical concepts

Multiple choice logarithm and its uses basic mathematical concepts physics

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

  1. $a < b$

  2. $ a > 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.

Multiple choice logarithm and its uses basic mathematical concepts physics

If $\log _{10}e=0.4343$, then $\log _{10}1016$ is

  1. $2.99$

  2. $3$

  3. $3.006949$

  4. $3.02$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$\log _{10}1016\Rightarrow \dfrac{\log 1016}{\log 10}$

$\Rightarrow \dfrac{3.006949}{1}$
$\rightarrow$ Option $C$ is correct

Multiple choice logarithm and its uses basic mathematical concepts physics

Multiple Correct:

Which of the following statements are true

  1. $\log _{ 2 }{ 3 } &lt;\log _{ 12 }{ 10 } $

  2. $\log _{ 6 }{ 5 } &lt;\log _{ 7 }{ 8 } $

  3. $\log _{ 3 }{ 26 } &lt;\log _{ 2 }{ 9 } $

  4. $\log _{ 16 }{ 15 } &gt;\log _{ 10 }{ 11 } &gt;\log _{ 7 }{ 6 }$
Reveal answer Fill a bubble to check yourself
B,C Correct answer
Multiple choice logarithm and its uses basic mathematical concepts physics

The solution of the equation $\log _{7}\log _{5}(\sqrt {x^{2}}+5+x)=0$

  1. $x=2$

  2. $x=3$

  3. $x=0$

  4. $x=-2$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$\log _{7}\log _{5}(\sqrt{x^{2}}+5+x)=0$

$\log _{5}(\sqrt{x^{2}}+5+x)=1$
$\sqrt{x^{2}+5+x=5}$
$\sqrt{x^{2}+x=0}$
$x^{2}+x^{2}+2\sqrt[x]{x^{2}}=0$
$x=0$


Multiple choice logarithm and its uses basic mathematical concepts physics

The value of $\displaystyle\sum _{r=1}^{n}log\left ( \dfrac{a^{r}}{b^{r-1}} \right )$ is

  1. $\dfrac{n}{2}log\left ( \dfrac{a^{n}}{b^{n}} \right )$

  2. $\dfrac{n}{2}log\left ( \dfrac{a^{n}}{b^{n+1}} \right )$

  3. $\dfrac{n}{2}log\left ( \dfrac{a^{n+1}}{b^{n+1}} \right )$

  4. $\dfrac{n}{2}log\left ( \dfrac{a^{n+1}}{b^{n-1}} \right )$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Now,
$\displaystyle\sum _{r=1}^{n}\log\left ( \dfrac{a^{r}}{b^{r-1}} \right )$
$=\displaystyle\sum _{r=1}^{n}\left(\log a^{r}-\log b^{r-1}\right)$
$=\displaystyle\sum _{r=1}^{n}\left(r\log a-(r-1)\log b\right)$
$=(\log a)\times \dfrac{n(n+1)}{2}-\log b\times\dfrac{(n-1)n}{2}$
$=\dfrac{n}{2}\left(\log a^{n+1}-\log b^{n-1}\right)$
$=\dfrac{n}{2}\log\left(\dfrac{a^{n+1}}{b^{n-1}}\right)$
Multiple choice logarithm and its uses basic mathematical concepts physics

Find the value of $\log _{10}{\left(0.\bar{9}\right)}$

  1. $0$

  2. $1$

  3. $-1$

  4. $2$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
To find value of $\log _{10}{0.\bar9}$

Let $x=0.\bar{9}=0.999999...$

$\Rightarrow 10x=9.99999....$

$\Rightarrow 10x-x=9$

$\Rightarrow 9x=9$

$\Rightarrow x=\dfrac{9}{9}=1$

$\therefore x=1$

Let $y=\log _{10}{1}$

$\Rightarrow 1={10}^{y}$

$\Rightarrow {10}^{y}={10}^{0}$                 (since ${10}^{0}=1$)

Since bases are same we can equate the powers

$\therefore y=0$

Hence, $\log _{10}{\left(0.\bar{9}\right)}=0$

Multiple choice logarithm and its uses basic mathematical concepts physics

If $y=a\log\left|x\right|+bx^{2}+x$ has extreme values at $x=2$ and $x=-4/3$ then 

  1. $a=12,b=-10$

  2. $a=4,b=-3/4$

  3. $a=-6,b=1/4$

  4. $none$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The derivative y' = a/x + 2bx + 1. Setting y' = 0 at x = 2 and x = -4/3 gives a system of equations: a/2 + 4b + 1 = 0 and a/(-4/3) - 8b/3 + 1 = 0. Solving these yields a=12 and b=-10.