Tag: law of indices

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Questions Related to law of indices

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

whether the following relation is${{ \frac{1}{{{x^{a - b}}}}} ^{\frac{1}{{a - c}}}}{{ \frac{1}{{{x^{b - c}}}}} ^{\frac{1}{{b - a}}}}{{ \frac{1}{{{x^{c - a}}}}} ^{^{\frac{1}{{c - b}}}}} = 1$

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using exponent rules, each term simplifies to x^(b-a)/(a-c) * x^(c-b)/(b-a) * x^(a-c)/(c-b). Adding the exponents (b-a)/(a-c) + (c-b)/(b-a) + (a-c)/(c-b) results in 0, and x^0 = 1.

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,

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

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

$x^2=1-1=0$

$\therefore x=0$