Tag: exponents

Quizzes (9) Questions (51)

Questions Related to exponents

Multiple choice comparison of irrational numbers surds and law of surds rational and irrational numbers exponents maths

Which among the following numbers is the greatest?
$\displaystyle \sqrt[3]{4},\sqrt{2},\sqrt[6]{13},\sqrt[4]{5}$

  1. $\displaystyle \sqrt[3]{4}$ is the greatest
  2. $\sqrt{2}$ is the greatest
  3. $\sqrt[6]{13}$ is the greatest
  4. $\sqrt[4]{5}$ is the greatest
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

LCM of $3, 6, 4 = 12$

So, raising each given number to power $12$.
$\Rightarrow \sqrt[3]{4}=(4)^{1/3}=(4^{1/3})^{12}=4^{4}=256$
$\Rightarrow \sqrt{2}=(2)^{1/2}=(2^{1/2})^{12}=2^{6}=64$
$\Rightarrow \sqrt[6]{13}=(13)^{1/6}=(13^{1/6})^{12}=13^{2}=169$
$\Rightarrow \sqrt[4]{5}=(5)^{1/4}=(15^{1/4})^{12}=5^{3}=125$

$\therefore \sqrt[3]{4}$ is the greatest.

Multiple choice comparison of irrational numbers surds and law of surds rational and irrational numbers exponents maths

If $A=\sqrt{7}-\sqrt{6}$ and $B=\sqrt{6}-\sqrt{5}$, then identify the true statement.

  1. $A> B$
  2. $A=B$
  3. $A< B$
  4. $A\ge B$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
$A=\sqrt{7}-\sqrt{6}\Rightarrow \dfrac{1}{A}=\dfrac{\sqrt{7}+\sqrt{6}}{(\sqrt{7}-\sqrt{6})(\sqrt{7}+\sqrt{6})}$
$\Rightarrow \boxed{\dfrac{1}{A}=\sqrt{7}+\sqrt{6}}$
$\boxed{\dfrac{1}{B}=\sqrt{6}+\sqrt{5}}$
As $\sqrt{7} > \sqrt{5}\Rightarrow \dfrac{1}{A} > \dfrac{1}{B}$
$\Rightarrow \boxed{B > A}$
Multiple choice comparison of irrational numbers surds and law of surds rational and irrational numbers exponents maths

The smallest between $\sqrt{17} - \sqrt{12}$ and $\sqrt{11} - \sqrt{6}$ is _________.

  1. $\sqrt{17} - \sqrt{12}$
  2. $\sqrt{11} - \sqrt{6}$

  3. Both are equal

  4. Can't be determined
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\sqrt{17} \approx 4.123$

$\sqrt{12} \approx 3.464$
$\sqrt{11} \approx 3.316$
$\sqrt{6} \approx 2.449$

$\Rightarrow \sqrt{17} - \sqrt{12} = 0.659$
$\Rightarrow \sqrt{11} - \sqrt{6} = 0.867$

Hence, $\sqrt{17}-\sqrt{12}$ is smaller.

Multiple choice comparison of irrational numbers surds and law of surds rational and irrational numbers exponents maths

The smallest of $\sqrt [ 3 ]{ 4 } , \sqrt [ 4 ]{ 5 } , \sqrt [ 4 ]{ 6 } , \sqrt [ 3 ]{ 8 } $ is:

  1. $\sqrt [ 3 ]{ 8 } $
  2. $\sqrt [ 4 ]{ 5 } $
  3. $\sqrt [ 3 ]{ 4 } $
  4. $\sqrt [ 4 ]{ 6 } $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
$\sqrt[3]{4}=\sqrt[12]{44}=\sqrt[12]{256}$
$\sqrt[4]{6}=\sqrt[12]{5^3}=\sqrt[12]{125}$
$\sqrt[4]{6}=\sqrt[12]{6^{3}}=\sqrt[12]{216}$
$\sqrt[3]{8}=\sqrt[12]{8^{4}}=\sqrt[12]{64^{2}}$
As $'125'$ is smallest
$\therefore \boxed{4\sqrt{5}}$ is smallest