ISEE Upper Level Quantitative Reasoning - How to find the exponent of variables

$t ^{3} = -24$

Evaluate $\left ( \frac{2}{t} \right ) ^{6}$ .

$\frac{1}{9}$

$-\frac{1}{9}$

$\frac{1}{72}$

$-\frac{1}{72}$

Explanation

By the Power of a Power Principle,

$(t ^{3} )^{2} = t ^{3 \cdot 2 } = t ^{6}$

By way of the Power of a Quotient Principle,

$\left ( \frac{2}{t} \right ) ^{6} = \frac{2^{6}}{t^{6} } = \frac{2^{6}}{(t^{3}) ^{2}} = \frac{64}{576} = \frac{64 \div 64 }{576 \div 64} = \frac{1}{9}$ .

is a negative number. Which is the greater quantity?

(a) $-16 a ^{4}$

(b) $(-2 a )^{4}$

(b) is the greater quantity

(a) is the greater quantity

(a) and (b) are equal

It is impossible to determine which is greater from the information given

Explanation

Any nonzero number raised to an even power, such as 4, is a positive number. Therefore,

$-16 a ^{4} = -16 \cdot a ^{4}$ is the product of a negative number and a positive number, and is therefore negative.

By the same reasoning, $(-2 a )^{4}$ is a positive number.

It follows that $(-2 a )^{4} > -16 a ^{4}$ .

Define $f(x) = x^{2}$ .

is a function with the set of all real numbers as its domain.

$(g \circ f)(5) = 60$

Which is the greater quantity?

(a)

(b)

It is impossible to determine which is greater from the information given

(b) is the greater quantity

(a) and (b) are equal

(a) is the greater quantity

Explanation

$f(x) = x^{2}$ , so $f(5) = 5^{2} = 25$ .

By definition,

$g[f(5)] = (g \circ f)(5)$ .

Since $(g \circ f)(5) = 60$ and , we can determine that

However, this does not tell us the value of at . Therefore, we do not know whether or , if either, is the greater.

Which is greater?

(a) $y ^{-2}$

(b) $y^{2}$

(b) is greater

(a) is greater

(a) and (b) are equal

It is impossible to tell from the information given

Explanation

If , then $y ^{2}> 1^{2} = 1$ and $y ^{-2}= \frac{1}{y ^{2}} < \frac{1}{1} = 1$

$y ^{2}> 1> y ^{-2}$ , so by transitivity, $y ^{2}> y ^{-2}$ , and (b) is greater

is a real number such that $t ^{6} = 121$ . Which is the greater quantity?

(a) $t^{3}$

(b) 11

It is impossible to determine which is greater from the information given

(a) is the greater quantity

(b) is the greater quantity

(a) and (b) are equal

Explanation

By the Power of a Power Principle,

$(t ^{3} ) ^{2} = t^{3 \cdot 2 } = t ^{6} = 121$

Therefore, $t ^{3}$ is a square root of 121, of which there are two - 11 and . Since it is possible for a third (odd-numbered) power of a real number to be positive or negative, we cannot eliminate either possibility, so either

Therefore, we cannot determine whether is less than 11 or equal to 11.

and are both real numbers.

$t^{4} = 121$

$u ^{4} = 81$

Evaluate .

Explanation

, as the product of a sum and a difference, can be rewritten using the difference of squares pattern:

$=(2t )^{2}-( 3u)^{2}$

$=2^{2} \cdot t ^{2}- 3^{2}\cdot u^{2}$

$=2^{2}t ^{2}- 3^{2}u^{2}$

$=4t ^{2}- 9u^{2}$

By the Power of a Power Principle,

$(t^{2} ) ^{2} = t^{2 \cdot 2 } = t^{4}$

Therefore, $t^{2}$ is a square root of $t^{4}$ - that is, a square root of 121. 121 has two square roots, and 121, but since is real, $t ^{2}$ must be the positive choice, 11. Similarly, $u^{2}$ is the positive square root of 81, which is 9.

The above expression can be evaluated as

$4t ^{2}- 9u^{2} = 4 (11) - 9 (9) = 44 - 81 = -37$ .

Simplify:

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

$\frac{8 y^{12} }{x^{9} }$

$\frac{8x^{9} }{ y^{12}}$

$\frac{1}{2y}$

$\frac{8 y^{7} }{x^{6} }$

$\frac{8x^{6} }{ y^{7}}$

Explanation

$\left (\frac{x^{3}}{2y^{4}} \right ) ^{-3} = \left (\frac{2y^{4}}{x^{3}} \right ) ^{3} = \frac{2^{3} \left (y^{4}\right ) ^{3} }{\left (x^{3} \right )^{3}}= \frac{8 y^{4 \cdot 3} }{x^{3\cdot 3} }= \frac{8 y^{12} }{x^{9} }$