GMAT Math : Understanding exponents

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Example Question #81 : Understanding Exponents

$\displaystyle A$ is the additive inverse of the multiplicative inverse of $\displaystyle B$; $\displaystyle C$ is the additive inverse of the multiplicative inverse of $\displaystyle D$. Which of the following is equal to the expression

$(AB)^{CD}$ $\displaystyle (AB)^{CD}$

regardless of the values of the variables?

Possible Answers:

$(AB)^{CD}$ $\displaystyle (AB)^{CD}$ is an undefined quantity

$\displaystyle 1$

$\displaystyle 2$

$\displaystyle -1$

$\displaystyle 0$

Correct answer:

$\displaystyle -1$

Explanation:

The additive inverse of a number is the number which, when added to that number, yields sum 0; the multiplicative inverse of a number is the number which, when multiplied by that number, yields product 1.

Let $\displaystyle X$ be the multiplicative inverse of $\displaystyle B$. Then

BX= 1 $\displaystyle BX= 1$, or, equivalently, $X = \frac{1}{B}$ $\displaystyle X = \frac{1}{B}$.

$\displaystyle A$ is the additive inverse of this number, so

$\displaystyle A + X = 0$

$A + \frac{1}{B}= 0$ $\displaystyle A + \frac{1}{B}= 0$

$A = - \frac{1}{B}$ $\displaystyle A = - \frac{1}{B}$

$A \cdot B = - \frac{1}{B} \cdot B$ $\displaystyle A \cdot B = - \frac{1}{B} \cdot B$

AB = -1 $\displaystyle AB = -1$

By similar reasoning, CD = -1 $\displaystyle CD = -1$, and

$(AB)^{CD} = \left ( -1\right ) ^{-1} = \frac{1}{-1} = -1$ $\displaystyle (AB)^{CD} = \left ( -1\right ) ^{-1} = \frac{1}{-1} = -1$

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Example Question #81 : Exponents

$FG= 2^{100}$ $\displaystyle FG= 2^{100}$

$HJ = 2^{10}$ $\displaystyle HJ = 2^{10}$

$\frac{F}{J} = 4^{40}$ $\displaystyle \frac{F}{J} = 4^{40}$

Which of the following is equal to $\frac{H}{G}$ $\displaystyle \frac{H}{G}$ ?

Possible Answers:

$\frac{1}{2^{170}}$ $\displaystyle \frac{1}{2^{170}}$

$\frac{1}{2^{50}}$ $\displaystyle \frac{1}{2^{50}}$

$\frac{1}{2^{10}}$ $\displaystyle \frac{1}{2^{10}}$

$\frac{1}{2^{130}}$ $\displaystyle \frac{1}{2^{130}}$

$\frac{1}{2^{70}}$ $\displaystyle \frac{1}{2^{70}}$

Correct answer:

$\frac{1}{2^{10}}$ $\displaystyle \frac{1}{2^{10}}$

Explanation:

Divide:

$\frac{HJ}{FG}=\frac{ 2^{10}}{2^{100}} = 2^{10-100} = 2^{-90}$ $\displaystyle \frac{HJ}{FG}=\frac{ 2^{10}}{2^{100}} = 2^{10-100} = 2^{-90}$

$\frac{HJ}{FG}= \frac{H}{G} \cdot \frac{J}{F} = \frac{H}{G} \div \frac{F} {J}$ $\displaystyle \frac{HJ}{FG}= \frac{H}{G} \cdot \frac{J}{F} = \frac{H}{G} \div \frac{F} {J}$

$\frac{F}{J} = 4^{40} = (2 ^{2})^{40} = 2 ^{2 \cdot 40 } = 2 ^{80}$ $\displaystyle \frac{F}{J} = 4^{40} = (2 ^{2})^{40} = 2 ^{2 \cdot 40 } = 2 ^{80}$

Substitute:

$\frac{H}{G} \div \frac{F} {J} = \frac{HJ}{FG}$ $\displaystyle \frac{H}{G} \div \frac{F} {J} = \frac{HJ}{FG}$

$\frac{H}{G} \div 2 ^{80}= 2^{-90}$ $\displaystyle \frac{H}{G} \div 2 ^{80}= 2^{-90}$

$\frac{H}{G} \div 2 ^{80}\cdot 2 ^{80} = 2^{-90} \cdot 2 ^{80}$ $\displaystyle \frac{H}{G} \div 2 ^{80}\cdot 2 ^{80} = 2^{-90} \cdot 2 ^{80}$

$\frac{H}{G} = 2^{-90+ 80} = 2^{-10} = \frac{1}{2^{10}}$ $\displaystyle \frac{H}{G} = 2^{-90+ 80} = 2^{-10} = \frac{1}{2^{10}}$

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Example Question #82 : Exponents

$\displaystyle A$ is the additive inverse of $\displaystyle B$; $\displaystyle C$ is the multiplicative inverse of $\displaystyle D$. Which of the following is equal to the expression

$(A+B)^{C+D}$ $\displaystyle (A+B)^{C+D}$

regardless of the values of the variables?

Possible Answers:

$A - \frac{1}{A}$ $\displaystyle A - \frac{1}{A}$

$\displaystyle 1$

$\displaystyle 0$

$A + \frac{1}{A}$ $\displaystyle A + \frac{1}{A}$

$(A+B)^{C+D}$ $\displaystyle (A+B)^{C+D}$ must be an undefined quantity.

Correct answer:

$\displaystyle 0$

Explanation:

The additive inverse of a number is the number which, when added to that number, yields sum 0. Since $\displaystyle A$ is the additive inverse of $\displaystyle B$,

$\displaystyle A + B = 0$

The multiplicative inverse of a number is the number which, when multiplied by that number, yields product 1. Since $\displaystyle C$ is the multiplicative inverse of $\displaystyle D$, then

CD = 1 $\displaystyle CD = 1$, or $D = \frac{1}{C}$ $\displaystyle D = \frac{1}{C}$.

It follows that

$(A+B)^{C+D} = 0^{C+ \frac{1}{C}}$ $\displaystyle (A+B)^{C+D} = 0^{C+ \frac{1}{C}}$.

0 raised to any nonzero power is equal to 0, and $C + \frac{1}{C}$ $\displaystyle C + \frac{1}{C}$ must be nonzero, so

$0^{C+ \frac{1}{C}} = 0$ $\displaystyle 0^{C+ \frac{1}{C}} = 0$, the correct response.

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Example Question #81 : Understanding Exponents

$\displaystyle A$ is the additive inverse of $\displaystyle B$; $\displaystyle C$ is the additive inverse of $\displaystyle D$. Which of the following is equal to the expression

$(A+B)^{C+D}$ $\displaystyle (A+B)^{C+D}$

regardless of the values of the variables?

Possible Answers:

$\displaystyle 0$

$A + \frac{1}{A}$ $\displaystyle A + \frac{1}{A}$

$\displaystyle 1$

$A - \frac{1}{A}$ $\displaystyle A - \frac{1}{A}$

$(A+B)^{C+D}$ $\displaystyle (A+B)^{C+D}$ must be an undefined quantity

Correct answer:

$(A+B)^{C+D}$ $\displaystyle (A+B)^{C+D}$ must be an undefined quantity

Explanation:

The additive inverse of a number is the number which, when added to that number, yields sum 0. Since $\displaystyle A$ is the additive inverse of $\displaystyle B$ and $\displaystyle C$ is the additive inverse of $\displaystyle D$,

$\displaystyle A + B = C+D = 0$

and

$(A+B)^{C+D} = 0^{0}$ $\displaystyle (A+B)^{C+D} = 0^{0}$,

which is an undefined expression.

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GMAT Math : Understanding exponents

Study concepts, example questions & explanations for GMAT Math

All GMAT Math Resources

Example Questions

Example Question #81 : Understanding Exponents

Example Question #81 : Exponents

Example Question #82 : Exponents

Example Question #81 : Understanding Exponents

All GMAT Math Resources

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