How to divide exponents - ISEE Upper Level Quantitative Reasoning

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Question

$\frac{ \left ( 30 -2 \cdot 15\right )^{0}}{\left ( 30 -15\right )^{0}}$

Answer

$\frac{ \left ( 30 -2 \cdot 15\right )^{0}}{\left ( 30 -15\right )^{0}} = \frac{ \left ( 30 -30\right )^{0}}{15^{0}}=\frac{ 0^{0}}{15^{0}}$

The numerator is undefined, since 0 raised to the power of 0 is an undefined quantity. Therefore, the entire expression is undefined.

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Question

Column A Column B

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

Answer

Let's simplify both quantities first before we compare them. $\left(\frac{1}{16}\right)^\frac{1}{2}$ becomes $\sqrt{\frac{1}{16}}$ because the fractional exponent indicates a square root. We can simplify that by knowing that we can take the square roots of both the numerator and denominator, as shown by: $\frac{\sqrt{1}}{\sqrt{16}}$ . We can simplify further by taking the square roots (they're perfect squares) and get $\frac{1}{4}$ . Then, let's simplify Column B. To get rid of the negative exponent, we put the numerical expression on the denominator. There's still the fractional exponent at play, so we'll have a square root as well. It looks like this now: $\frac{1}{\sqrt{\frac{1}{16}}}$ . We already simplified $\sqrt{\frac{1}{16}}$ , so we can just plug in our answer, $\frac{1}{4}$ , into the denominator. Since we don't want a fraction in the denominator, we can multiply by the reciprocal of $\frac{1}{4}$ , which is 4 to get , which is just 4. Therefore, Column B is greater.

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Question

Column A Column B

$\frac{x^3}{x^2}$ $\sqrt{x}$

Answer

You can simplify Column A first. When you're dividing with exponents and bases are the same, subtract the exponents. Therefore, it simplifies to x. We know that x is positive since it is greater than 1. X is greater than $\sqrt{x}$ . Try plugging in a number to test. 25 is greater than $\sqrt{25}$ , which is 5. Even 1.1 is greater than $\sqrt{1.1}$ . Therefore, Column A is greater.

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Question

Give the reciprocal of $1.6 \times 10 ^{-7}$ in scientific notation.

Answer

The reciprocal of $1.6 \times 10 ^{-7}$ is the quotient of 1 and the number;

$\frac{1}{ 1.6 \times 10 ^{-7} }$

$= \frac{1 \times 10^{0}}{ 1.6 \times 10 ^{-7} }$

$= \frac{1 }{ 1.6 } \times \frac{ 10^{0}}{ 10 ^{-7} }$

$=0.625 \times 10^{0- (-7) }$

$=0.625 \times 10^{7 }$

This is not in scientific notation, so adjust.

$6.25 \times 10^{-1 } \times 10^{7 }$

$= 6.25 \times 10^{-1+7 }$

$= 6.25 \times 10^{6 }$

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Question

Give the reciprocal of $3.2 \times 10 ^{9}$ in scientific notation.

Answer

The reciprocal of $3.2 \times 10 ^{9}$ is the quotient of 1 and the number, or

$\frac{1}{ 3.2 \times 10 ^{9} }$

$= \frac{1 \times 10^{0}}{ 3.2 \times 10 ^{9}}$

$= \frac{1 }{ 3.2 } \times \frac{ 10^{0}}{ 10 ^{9} }$

$=0.3125 \times 10 ^{-9}$

This is not in scientific notation, so adjust:

$3.125 \times 10 ^{-1}\times 10 ^{-9}$

$=3.125 \times 10 ^{-1+ (-9)}$

$=3.125 \times 10 ^{-10}$

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Question

Simplify: $\frac{x^{50}}{x^{10}}$

Answer

Apply the quotient of powers rule.

$\frac{x^{50}}{x^{10}} = x^{50-10} = x^{40}$

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Question

Simplify: $\frac{12x^{12}}{3x^{3}}$

Answer

Separate the fraction and apply the quotient of powers rule:

$\frac{12x^{12}}{3x^{3}} = \frac{12}{3} \cdot \frac{x^{12}}{x^{3}} = 4x^{12-3} = 4x^{9}$

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Question

Evaluate:

$\frac{120}{12^{0}}$

Answer

Any nonzero number raised to the power of 0 is equal to 1, so $12^{0} = 1$ , and

$\frac{120}{12^{0}} = \frac{120}{1} = 120$

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Question

Evaluate:

$\frac{25}{10 ^{0} -5^{0}}$

Answer

Any nonzero number raised to the power of 0 is equal to 1, so $5^{0} = 10 ^{0} =1$ . Therefore,

$\frac{25}{10 ^{0} -5^{0}} = \frac{25}{1-1} = \frac{25}{0}$

However, an expression with a denominator of 0 is undefined, so that is the correct choice.

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Question

Simplify:

$\frac{6x^6}{2x^3}$

Answer

Based on the quotient rule for exponents, we know that for any non-zero number and any integers and ,

$\frac{x^a}{x^b}=x^{a-b}$ .

We should separate the fraction and apply the quotient rule for exponents:

$\frac{6x^6}{2x^3}=\frac{6}{2}\times \frac{x^6}{x^3}=3\times x^{6-3}$

$\Rightarrow \frac{6x^6}{2x^3}=3x^3$

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Question

Simplify:

$\frac{4x^8}{2x^0}$

Answer

Based on the quotient rule for exponents, we know that for any non-zero number and any integers and ,

$\frac{x^a}{x^b}=x^{a-b}$ .

We also know that any non-zero number raised to the power of is .

$\frac{4x^8}{2x^0}=\frac{4}{2}\times \frac{x^8}{1}=2x^8$

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Question

Simplify:

$\frac{6x^4-6}{3x^2+3x^0}$

Answer

We know that any non-zero number raised to the power of is equal to .

$\frac{6x^4-6}{3x^2+3x^0}=\frac{6(x^4-1)}{3(x^2+x^0)}=\frac{6(x^4-1)}{3(x^2+1)}$

Now factor:

$\frac{6(x^4-1)}{3(x^2+1)}=\frac{6}{3}\frac{(x^2-1)(x^2+1)}{x^2+1}=2(x^2-1)$

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Question

Evaluate:

$\frac{4^0}{2\times 5^0-12^0}$

Answer

We know that any non-zero number raised to the power of is equal to .

$\frac{4^0}{2\times 5^0-12^0}=\frac{1}{2\times 1-1}=\frac{1}{2-1}=1$

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Question

Evaluate:

$\frac{y^{-2}}{y^{-5}}$

Answer

Based on the quotient rule for exponents, we know that, for any non-zero number and any integers and ,

$\frac{x^a}{x^b}=x^{a-b}$ .

$\frac{y^{-2}}{y^{-5}}=y^{-2-(-5)}=y^{-2+5}=y^{3}$

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Question

Simplify:

$(\frac{3x}{y^3})^{3}$

Answer

When a fraction is raised to a power, the numerator and the denominator are both raised to that power:

$(\frac{3x}{y^3})^{3}=\frac{(3x)^3}{(y^3)^3}$

The power rule says that in order to raise a power to a power, we need to multiply the exponents, i.e.

$(x^{m})^{n}=x^{m\times n}$ ,

where and are real numbers.

$(\frac{3x}{y^3})^{3}=\frac{(3x)^3}{(y^3)^3}=\frac{(3^{3}\times x^3)}{y^{3\times 3}}=\frac{27x^3}{y^9}$

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Question

Evaluate:

$(\frac{x}{y^4})^{5}$

Answer

When a fraction is raised to a power, the numerator and the denominator are both raised to that power:

$(\frac{x}{y^4})^{5}=\frac{x^5}{(y^4)^5}$

The power rule says that in order to raise a power to a power, we need to multiply the exponents, i.e.

$(x^{m})^{n}=x^{m\times n}$ .

$(\frac{x}{y^4})^{5}=\frac{x^5}{(y^4)^5}=\frac{x^5}{y^{4\times 5}}=\frac{x^5}{y^{20}}$

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Question

Simplify:

$\frac{x^{-4}}{y^{-8}}$

Answer

The Negative Exponent Rule says $a^{-n}=\frac{1}{a^n}$ .

$\frac{x^{-4}}{y^{-8}}=\frac{y^8}{x^4}$

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Question

Evaluate:

$(\frac{14x^2}{7x})^0$

Assume variables are non-zero.

Answer

We know that any non-zero number raised to the power of is equal to .

$(\frac{14x^2}{7x})^0=1$

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Question

Evaluate:

$(\frac{22x^3}{11y})^5$

Answer

When a fraction is raised to a power, the numerator and the denominator are both raised to that power. In addition, the power rule says that in order to raise a power to a power, multiply the exponents, i.e.

$(x^{m})^{n}=x^{m\times n}$ .

$(\frac{22x^3}{11y})^{5}=\frac{(22x^3)^{5}}{(11y)^5}=(\frac{22}{11})^5\times \frac{x^{3\times 5}}{y^5}=2^5\times \frac{x^{15}}{y^5}=\frac{32x^{15}}{y^5}$

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Question

Simplify:

$\frac{(2x^4y^{-1})^{-3}}{(3x^3y^2)^{-2}}$

Answer

Based on the negative exponent rule we have:

$a^{-n}=\frac{1}{a^n}$

Which says negative exponents in the numerator get moved to the denominator and become positive exponents. And negative exponents in the denominator get moved to the numerator and become positive exponents. So we can write:

$\frac{(2x^4y^{-1})^{-3}}{(3x^3y^2)^{-2}}=\frac{(3x^3y^2)^{2}}{(2x^4y^{-1})^{3}}$

Based on the power rule for exponents we can write:

$(a^m)^{n}=a^{m\times n}$

That means; to raise a power to a power we need to multiply the exponents.So we have:

$\frac{(3x^3y^2)^{2}}{(2x^4y^{-1})^{3}}=\frac{3^2\times x^{(3\times 2)}\times y^{(2\times 2)}}{2^3\times x^{(4\times 3)}\times y^{(-1\times 3)}}=\frac{9x^6y^4}{8x^{12}y^{-3}}$

Separate the equation:

$\Rightarrow \frac{9x^6y^4}{8x^{12}y^{-3}}=\frac{9}{8}\times \frac{x^6}{x^{12}}\times \frac{y^4}{y^{-3}}$

$=\frac{9}{8}\times x^{6-12}\times y^{4+3}=\frac{9}{8}x^{-6}y^7=\frac{9y^7}{8x^6}$

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