Tag: power of 10

Questions Related to power of 10

If $4^{n-2} + 4^{2} = 32$, then what is the value of $n$?

  1. $2$

  2. $4$

  3. $6$

  4. $8$


Correct Option: B
Explanation:

Given, ${ 4 }^{ n-2 }+{ 4 }^{ 2 }=32$
${ 4 }^{ n-2 }=32-{ 4 }^{ 2 }$
${ 4 }^{ n-2 }=32-16$
${ 4 }^{ n-2 }=16$
${ 4 }^{ n-2 }={ 4 }^{ 2 }$
$n-2=2$ ..... (as the bases are equal)
$n=4$

If $\sqrt { { 2 }^{ x } } =16$, then $x=$

  1. $8$

  2. $4$

  3. $2$

  4. $10$


Correct Option: A
Explanation:

$\sqrt { { 2 }^{ x } } =16\ \Rightarrow { (2) }^{ \dfrac { x }{ 2 }  }={ 2 }^{ 4 }$


Exponents should be equal.

$ \dfrac { x }{ 2 } =4\Rightarrow x=8$

If $2^x - {2^{x - 1}} = 4$ then $x^x$ is equals to 

  1. $7$

  2. $3$

  3. $27$

  4. None of these


Correct Option: C
Explanation:
$2^{x}-2^{x-1}=4$

$2^{x}-2^{x}2^{-1}=4$

$2^{x}\left (1-\dfrac{1}{2}\right )=2^{x-1}=4=2^2$

$\therefore$ $x-1=2$

$\text{from this we get }$  $x=3$

$x^{x}=3^{3}=27$

If $3^x=5^y=7^z=105$, then $\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}$ is equal to?

  1. $1$

  2. $0$

  3. $-1$

  4. None


Correct Option: A
Explanation:

$x = \cfrac{\log 105}{\log 3}, y = \cfrac{\log 105}{\log 5} , z = \cfrac{\log 105}{\log 7}$

$\cfrac{1}{x} + \cfrac{1}{y} + \cfrac{1}{z} = \cfrac{\log 3 + \log 5 + \log 7}{\log 105} = \cfrac{\log 105}{\log 105} = 1$

Which is greatest among following $2^{156},\ 4^{79},\ 128^{23}$ and $8^{54}$?

  1. $4^{79}$

  2. $128^{23}$

  3. $2^{156}$

  4. $8^{54}$


Correct Option: D
Explanation:
$2^{156}\rightarrow 2^{156}$
$4^{79}\rightarrow {2^{2\times 79}}\rightarrow 2^{158}$
$128^{23}\rightarrow 2^{7\times 23}\rightarrow 2^{161}$
$8^{54}\rightarrow 2^{3\times 54}\rightarrow 2^{162}$
$8^{54}$ is greatest
D is correct.

If $ { 9 }^{ x-1 }={ 3 }^{ 2x-1 }-486 $,then the value of x is:

  1. $\dfrac{7}{2}$

  2. 4

  3. 1

  4. 0


Correct Option: A
Explanation:
$9^{x-1}=3^{2x-1}-486$
$3^{2x-1}=9^{x-1}t486$
$\dfrac{3^{2x}}{3}=3^{2x-2}+486$
$3^{2x}=\dfrac{3(3^{2x})}{3^{z}}t(486)3$
$3^{2q}=3^{2x-1}+1458$
$2177=729+1458$
$3^{7}=3^{6}+1458$
$2x=7$
$x=\dfrac{7}{2}$

$For \ a \  natural \  number \  n , \  2n(n-1)!\leqslant  n^n, if$

  1. $n<2$

  2. $n>2$

  3. $n\ge2$

  4. Never


Correct Option: C

Which of the following represents the given expression?
$a^2b^3\times 2ab^2$ ?

  1. $2a^3b^4$

  2. $2a^3b^5$

  3. $2ab$

  4. $a^3b^5$


Correct Option: B
Explanation:
We know that if a number say $b$ is multiplied three times. That is, $b\times b\times b$ can be written as $b^3$.
=> $b^{3}$ = $b\times b\times b$
Similarly,
$a^{2}b^3$ = $a \times a\times b\times b\times b$
$2ab^{2}$ = $ 2 \times a\times b\times b$.
Thus, 
$a^{2}b^{3}\times 2ab^{2} = a \times a\times b\times b\times b \times 2 \times a\times b\times b $.
                      $= 2 \times a \times a \times a\times b\times b\times b\times b\times b$.
                      $ = 2a^{3}b^{5}$.

If $\displaystyle 2^{2^{3}}=j, 2^{3^{2}}=k, 3^{2^{2}}=\varphi ,$ then 

  1. $k = 2j$

  2. $j < k$

  3. $\displaystyle \varphi < k$

  4. All of these


Correct Option: D
Explanation:

  $j=2^{2{^3}}$, $k=2^{3^{2}}$, $\varphi=3^{2^{2}}$

$\Rightarrow$  $j=256,\,\,k=512,\,\,\varphi=81$
$\Rightarrow$  Option A says $k=2j$, which is true because $j=256$, so $k=2\times 256=512$
$\Rightarrow$  Option B says $j<k$, which is true because value of $j$ is $255$ and value of $k$ is $512$
$\Rightarrow$   Option C says $\varphi <k$, which is true because value of $\varphi$ is $81$ and value of $k$ is $512$

If $\displaystyle a^{m}=b^{m}$ and $(m > 0)$, then which of the following options could be true:

  1. $a = -b$

  2. $a  + b= 0$

  3. $2a - b = 0$

  4. $\displaystyle \dfrac{a^{2}}{b^{2}}=1$


Correct Option: D
Explanation:

Given $a^m=b^m$


Dividing by $b^m$

$\dfrac{a^m}{b^m}=1$
${\left(\dfrac{a}{b}\right)}^m=1$

Only Option D satisfies this answer.