Tag: maths

Quizzes (518) Questions (9324)

Questions Related to maths

Multiple choice maths power and exponent power of powers laws of exponents and powers law of indices

The sum of roots of the equation $(1.25)^{1-x^2} = (0.4096)^{1+x}$

  1. Infinite
  2. $1$
  3. $2$
  4. $4$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
$\left ( 1.25 \right )^{1-x^{2}}=\left ( 0.4096 \right )^{1+x}$

$\left ( \dfrac{125}{100} \right )^{1-x^{2}}=\left ( \dfrac{4096}{10000} \right )^{1+x}$

$\left ( \dfrac{5}{4} \right )^{1-x^{2}}=\left ( \left ( \dfrac{8}{10} \right )^{4} \right )^{1+x}$

$\left ( \dfrac{5}{4} \right )^{1-x^{2}}=\left ( \dfrac{4}{5} \right )^{4+4x}$

$\left ( \dfrac{5}{4} \right )^{1-x^{2}}=\left ( \dfrac{5}{4} \right )^{-4-4x}$

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

$x^{2}-4x-5=0$

$x^{2}-5x+x-5=0$
$x(x-5)+1(x-5)=0$
$(x+1)(x-5)=0$
$x=-1,5$

Therefore, Sum of the roots of equation  is $4$
Multiple choice maths power and exponent power of powers laws of exponents and powers law of indices

Find:$\dfrac{\sqrt[3]{108}\times \sqrt[6]{4}}{\sqrt[4]{81}}$

  1. $ 2$
  2. $\frac{\sqrt[6]{4}}{\sqrt[2]{3}}$
  3. $ \sqrt[4]{6}$
  4. $\frac{\sqrt[4]{2}}{\sqrt[2]{3}}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$(\frac{\sqrt[3]{180}\times \sqrt[6]{4}}{\sqrt[4]{81}})\\ ((\frac{27\times4)^{(\frac{1}{4})}\times 2^{(\frac{2}{6})}}{(3^4)^{(\frac{1}{4})}}))\\ =(\frac{3\times 2^{(\frac{2}{3}) }\times 2^{(\frac{1}{3})}}{3})\\= 2^{{(\frac{2}{3})}+{(\frac{1}{3})}}\\=2$

Multiple choice maths power and exponent power of powers laws of exponents and powers law of indices

If $ p= {2} ^{ \tfrac {2} {3}} + {2} ^{ \tfrac {1} {3}} $,then 

  1. ${p}^{3}-6p+6=0 $
  2. ${p}^{3}-3p-6=0 $
  3. ${p}^{3}-6p-6=0 $
  4. ${p}^{3}-3p+6=0 $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Given,

$p=2^{\frac{2}{3}}+2^{\frac{1}{3}}$

cubing both sides

$p^3=(2^{\frac{2}{3}}+2^{\frac{1}{3}})^3$

$p^3=\left(2^{\frac{2}{3}}\right)^3+3\left(2^{\frac{2}{3}}\right)^2\cdot \:2^{\frac{1}{3}}+3\cdot \:2^{\frac{2}{3}}\left(2^{\frac{1}{3}}\right)^2+\left(2^{\frac{1}{3}}\right)^3$

$p^3=6+6\cdot \:2^{\frac{2}{3}}+6\cdot \:2^{\frac{1}{3}}\quad $

$p^3=6+6\left ( 2^{\frac{2}{3}}+2^{\frac{1}{3}} \right )$

$p^3=6+6p$

$\therefore p^3-6-6p=0$
Multiple choice maths power and exponent power of powers laws of exponents and powers law of indices

$\dfrac{(625)^{6.25} \times (25)^{2.6}}{(625)^{6.75} \times (5)^{1.2}} = ?$

  1. $5$
  2. $10$
  3. $15$
  4. $25$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Given,

$\dfrac{625^{6.25}\cdot \:25^{2.6}}{625^{6.75}\cdot \:5^{1.2}}$

$=\dfrac{625^{\tfrac{25}{4}}\cdot \:25^{\tfrac{13}{5}}}{625^{\tfrac{27}{4}}\cdot \:5^{\tfrac{6}{5}}}$

$=\dfrac{(5^4)^{\tfrac{25}{4}}\cdot \:(5^2)^{\tfrac{13}{5}}}{(5^4)^{\tfrac{27}{4}}\cdot \:5^{\tfrac{6}{5}}}$

$=\dfrac{(5^{25})\cdot \:(5)^{\tfrac{26}{5}}}{(5)^{27}\cdot \:5^{\tfrac{6}{5}}}$

$=\dfrac{5^{25+\tfrac{26}{5}}}{5^{27+\tfrac{6}{5}}}$

$=5^{25+\tfrac{26}{5}-27-\tfrac{6}{5}}$

$=5^{-2+\tfrac{20}{5}}$

$=5^{-2+4}$

$=5^2=25$
Multiple choice maths power and exponent power of powers laws of exponents and powers law of indices

Find $x:[3+\left { 2+(1+x^{2}) \right }^{2}]^{2}=144$

  1. $1$
  2. $0$
  3. $5$
  4. $6$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Given,

$\left [ 3+\left\{2+\left(1+x^2\right)\right\}^2 \right ]^2=144$

taking square root on both sides, we get,

$\left [ 3+\left\{2+\left(1+x^2\right)\right\}^2 \right ]=12$

$\left\{2+\left(1+x^2\right)\right\}^2=12-3=9$

again taking square root on both sides, we get,

$2+(1+x^2)=3$

$1+x^2=3-2=1$

$x^2=1-1=0$

$\therefore x=0$
Multiple choice maths power and exponent power of powers laws of exponents and powers law of indices

Simplicity
$\left[ \left{ \left( 625 \right) ^{ -\dfrac { 1 }{ 2 } } \right} ^{ -\dfrac { 1 }{ 4 } } \right] $

  1. $\dfrac{1}{\sqrt5}$
  2. $\sqrt5$

  3. 5

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

$\begin{array}{l}\left[ {{{\left( {{{\left( {625} \right)}^{\frac{{ - 1}}{2}}}} \right)}^{\frac{{ - 1}}{4}}}} \right] = \left[ {{{\left( {{{\left( {{{25}^2}} \right)}^{\frac{{ - 1}}{2}}}} \right)}^{\frac{{ - 1}}{4}}}} \right]\ = \left[ {{{\left( {{{25}^{ - 1}}} \right)}^{\frac{{ - 1}}{4}}}} \right]\ = {25^{\frac{1}{4}}}\ = {5^{2 \times \frac{1}{4}}}\ = {5^{\frac{1}{2}}}\ = \sqrt 5 \end{array}$