Understanding powers and roots - GMAT Quantitative

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Question

Evaluate $\left (1,296^{\frac{1}{3} } \right )^{\frac{1}{2} }$ .

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Answer

$1,296 = 6 ^{4}$ , so

$\left (1,296^{\frac{1}{3} } \right )^{\frac{1}{2} } = \left [ \left (6 ^{4 } \right )^{\frac{1}{3} } \right ]^{\frac{1}{2} } = 6 ^{4 \cdot \frac{1}{3} \cdot \frac{1}{2}} =6 ^{ \frac{2}{3} }$

For all integers and all positive bases ,

$a^{ \frac{M}{N}} = \sqrt[N]{a^{M}}$ by definition. So set

:

$\left (1,296^{\frac{1}{3} } \right )^{\frac{1}{2} }=6 ^{ \frac{2}{3} } = \sqrt[3]{6 ^{2}} = \sqrt[3]{36}$

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Question

Simplify by rationalizing the denominator:

$\frac{1+\sqrt{2}}{1+\sqrt{3}}$

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Answer

Multiply both numerator and denominator by the conjugate of the denominator, which is $1-\sqrt{3}$ :

$\frac{1+\sqrt{2}}{1+\sqrt{3}}$

$=\frac{(1+\sqrt{2})(1-\sqrt{3})}{(1+\sqrt{3})(1-\sqrt{3})}$

$=\frac{ 1 \cdot 1-1 \cdot \sqrt{3}+ \sqrt{2} \cdot 1-\sqrt{2} \cdot \sqrt{3}} { 1^{2}-(\sqrt{3})^{2}}$

$=\frac{1- \sqrt{3}+ \sqrt{2} -\sqrt{6} } { 1-3}$

$=\frac{1+ \sqrt{2} - \sqrt{3}-\sqrt{6} } { -2}$

$=\frac{-1- \sqrt{2} + \sqrt{3}+\sqrt{6} } { 2}$

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Question

$(\sqrt{75}+\sqrt{108})^{\sqrt[3]{8}}=$

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Answer

$(\sqrt{75}+\sqrt{108})^{\sqrt[3]{8}}=(\sqrt{75}+\sqrt{108})^{2}$

$=(\sqrt{75})^{2}+(\sqrt{108})^{2}+2\times \sqrt{75}\times \sqrt{108}$

$=75+108+2\times \sqrt{3(5^{2})}\times \sqrt{3(6^{^{2}})}$

$=75+108+2\times5\sqrt{3}\times6\sqrt{3}$

$=183+60(\sqrt{3})^{2}$

$=183+60\times3$

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Question

Which of the following numbers is irrational?

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Answer

A square root of a number is irrational if the number inside the square root is not a perfect square. Since 2 is not a perfect square, $\sqrt{2}$ is irrational

$\sqrt[3]{8}$ is rational, because is a perfect cube. ( $2\times2\times2 = 8$ )

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Question

What is ?

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Answer

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Question

What is $\sqrt{81}$ ?

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Answer

$\sqrt{81}=9$

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Question

Evaluate $24 ^{\frac{2}{3}}$ , expressing the result as a simplified radical if applicable.

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Answer

For all integers and all positive bases ,

$a^{ \frac{M}{N}} = \sqrt[N]{a^{M}}$ by definition.

Set :

$24^{ \frac{2}{3}} = \sqrt[3]{24^{2}} = \sqrt[3]{576} = \sqrt[3]{64} \cdot \sqrt[3]{9}= 4 \sqrt[3]{9}$

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Question

Which of the following is equivalent to $\sqrt[3]{\sqrt[4]{x}}$ ?

You may assume to be positive.

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Answer

Convert the roots to fractional exponents, then back, as follows:

$\sqrt[3]{\sqrt[4]{x}} = \sqrt[3]{ x^{\frac{1}{4}}} = \left ( x^{\frac{1}{4}} \right ) ^{\frac{1}{3}}= x ^{\frac{1}{4} \cdot \frac{1}{3}} = x^{\frac{1}{12}}= \sqrt[12]{x}$

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Question

Evaluate $\left (625^{\frac{1}{2} } \right )^{\frac{1}{2} }$ .

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Answer

$\left (625^{\frac{1}{2} } \right )^{\frac{1}{2} } =\left ( \sqrt{625}\right )^{\frac{1}{2} } = \left ( 25\right )^{\frac{1}{2} }= \sqrt{25} = 5$

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Question

What is $\sqrt{225}$ ?

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Answer

$\sqrt{225}=15$

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Question

Solve: $\small (\sqrt{5}+\sqrt{4})^2$

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Answer

First, FOIL:

$\small (\sqrt5+\sqrt4)^2=(\sqrt5+\sqrt4)(\sqrt5+\sqrt4)=5+\sqrt20+\sqrt20+4$

$\small =9+2\sqrt20$

Factor out $\small \sqrt4$

$\small =9+4\sqrt5$

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Question

Solve: $\small \frac{10^9-10^7}{99}$

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Answer

First factor. $\small \frac{10^9-10^7}{99}\ =\ \frac{(10^7)(10^2-1)}{99}$

Simplify. $\small \frac{(10^7)(10^2-1)}{99}\ =\ \frac{(10^7)(100-1)}{99}\ =\ \frac{(10^7)(99)}{99}\ =\ 10^7$

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Question

If $\small x^2=8$ , what is $\small x^4?$

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Answer

$\small x^4=(x^2)^2=(8)^2=64$

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Question

Evaluate $\left (-256^{\frac{1}{2} } \right )^{\frac{1}{2} }$ .

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Answer

$256 ^{\frac{1}{2}} = \sqrt{256} = 16$ , so

$\left (-256^{\frac{1}{2} } \right )^{\frac{1}{2} } = \left (-16\right )^{\frac{1}{2} } = \sqrt{-16}$

However, has no real square root. The expression is not equal to a real number.

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Question

If the side length of a cube is tripled, how does the volume of the cube change?

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Answer

The equation for the volume of a cube is $L^{3}$ . If the length is tripled, it becomes $(3L)^{3}$ , and $3^{3}=27$ , so the volume increases by 27 times the original size.

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Question

Evaluate: $11^{100} - 11^{99}$

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Answer

$11^{100} - 11^{99} = 11 \cdot 11^{99} - 1\cdot 11^{99} = (11-1) \cdot 11^{99} = 10 \cdot 11^{99}$

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Question

Solve: $\frac{(0.5)^{6}}{(0.5)^{9}}$

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Answer

Solve $\dpi{100} \small : \frac{0.5^{6}}{0.5^{9}} = 0.5^{6-9}= 0.5^{-3}=\frac{1}{0.5^{3}}=\frac{1}{0.125} = 8$

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Question

$\frac{x^{2}y^{3}z^{4}}{x^{3}yz^{3}} =$

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Answer

$\frac{x^{2}y^{3}z^{4}}{x^{3}yz^{3}} = \frac{y^{3-1}z^{4-3}}{x^{3-2}} = \frac{y^{2}z}{x}$

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Question

In the sequence 1, 3, 9, 27, 81, … , each term after the first is three times the previous term. What is the sum of the 9th and 10th terms in the sequence?

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Answer

We can rewrite the sequence as $(3)^{0}$ , $(3)^{1}$ , $(3)^{2}$ , $(3)^{3}$ , $(3)^{4}$ , … ,

and we can see that the 9th term in the sequence is $(3)^{8}$ and the 10th term in the sequence is $(3)^{9}$ . Therefore, the sum of the 9th and 10th terms would be

$3^{8}+3^{9}=3^{8}\cdot \left ( 1+3 \right )=4\cdot 3^{8}$

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Question

Simplify this expression as much as possible:

$\sqrt{9x}+\sqrt{12x}+\sqrt{27x}$

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Answer

$\sqrt{9x}+\sqrt{12x}+\sqrt{27x}$

$=\sqrt{9}\cdot \sqrt{x}+\sqrt{4}\cdot \sqrt{3x}+\sqrt{9}\cdot \sqrt{3x}$

$=3 \sqrt{x}+2 \sqrt{3x}+3 \sqrt{3x}$

$=3 \sqrt{x}+\left (2 +3 \right )\sqrt{3x}$

$=3 \sqrt{x}+5\sqrt{3x}$

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