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

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Question

Simplify:

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

Answer

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

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Question

Which is greater?

(a) $y ^{-2}$

(b) $y^{2}$

Answer

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

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Question

Which is greater?

(a) $\left (-2 \right )^{-3}$

(b) $\left ( \frac{1}{2} \right ) ^{3}$

Answer

$\left (-2 \right )^{-3} = \left ( -\frac{1}{2} \right )^{3}$

A negative number to an odd power is negative, so the expression in (a) is negative. The expression in (b) is positive since the base is positive. (b) is greater.

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Question

Expand: $(x + 1)^{10}$

Which is the greater quantity?

(a) The coefficient of $x^{3}$

(b) The coefficient of $x^{7}$

Answer

By the Binomial Theorem, if $(x + 1)^{n}$ is expanded, the coefficient of $x^{r}$ is

$_{n}\textrm{C} _{r} = \frac{n!}{(n-r)!r!}$ .

(a) Substitute : The coerfficient of $x^{3}$ is

$_{10}\textrm{C} _{3} = \frac{10!}{(10-3)!3!} = \frac{10!}{7!3!}$ .

(b) Substitute : The coerfficient of $x^{7}$ is

$_{10}\textrm{C} _{7} = \frac{10!}{(10-7)!7!} = \frac{10!}{3!7!} = \frac{10!}{7!3!}$ .

The two are equal.

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Question

Expand: $(x + 2)^{10}$

Which is the greater quantity?

(a) The coefficient of

(b) The coefficient of $x^{2}$

Answer

Using the Binomial Theorem, if $(x + 2)^{n}$ is expanded, the $x^{r}$ term is

$_{n}\textrm{C} _{r} \cdot x^{r} \cdot 2 ^{n-r}$

$= \frac{n!}{(n-r)!r!} \cdot 2 ^{n-r} \cdot x^{r}$ .

This makes $\frac{n!}{(n-r)!r!} \cdot 2 ^{n-r}$ the coefficient of $x^{r}$ .

We compare the values of this expression at for both and .

(a) If and , the coefficient is

$\frac{10!}{(10-1)!1!} \cdot 2 ^{10-1} = \frac{10!}{9!} \cdot 2 ^{9} =10 \cdot 512 =5,120$ .

This is the coefficient of .

(b) If and , the coefficient is

$\frac{10!}{(10-2)!2!} \cdot 2 ^{10-2} = \frac{10!}{8!2!} \cdot 2 ^{8} = \frac{ 9 \cdot 10}{2} \cdot 256 =11,520$ .

This is the coefficient of $x^{2}$ .

(b) is the greater quantity.

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Question

Which is the greater quantity?

(a) $(x + 5)^{3}- (x - 5)^{3}$

(b)

Answer

Simplify the expression in (a):

$(x + 5)^{3}- (x - 5)^{3}$

$=(x^{3}+ 3 \cdot x^{2} \cdot 5 + 3 \cdot x \cdot 5^{2}+ 5^{3}) -(x^{3}- 3 \cdotx^{2} \cdot 5 + 3 \cdot x \cdot 5^{2}- 5^{3})$

$=(x^{3}+15 x^{2}+75 x + 125) - (x^{3}-15 x^{2}+75 x- 125)$

$=x^{3}-x^{3}+15 x^{2}+15 x^{2}+75 x-75 x + 125+125$

$=30 x^{2} + 250$

Since ,

$(x + 5)^{3}- (x - 5)^{3}=30 x^{2} + 250 > 250$ ,

making (a) greater.

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Question

Consider the expression $\left (x^{2} y \right )^{3} \left (x y ^{2} \right )^{3}$

Which is the greater quantity?

(a) The expression evaluated at

(b) The expression evaluated at

Answer

Use the properties of powers to simplify the expression:

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

$= \left (x^{2} \right )^{3} y ^{3} \cdot x ^{3}\left ( y ^{2} \right )^{3}$

$= \left (x^{2} \right )^{3} \cdot x ^{3}\cdot y ^{3} \cdot \left ( y ^{2} \right )^{3}$

$= \left (x^{2 \cdot 3} \right ) \cdot x ^{3}\cdot y ^{3} \cdot \left ( y ^{2\cdot 3} \right )$

$= x^{6} \cdot x ^{3}\cdot y ^{3} \cdot y ^{6}$

$= x^{6+3} \cdot y ^{3+ 6}$

$= x^{9} y ^{9} = (xy)^{9}$

(a) If , then

$\left (x^{2} y \right )^{3} \left (x y ^{2} \right )^{3}= (xy)^{9} = ( 6 \cdot 10)^{9} = 60^{9}$

(b) If , then

$\left (x^{2} y \right )^{3} \left (x y ^{2} \right )^{3}= (xy)^{9} = ( 8 \cdot 8)^{9} = 64^{9}$

(b) is greater.

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Question

Which of the following expressions is equivalent to

$\left (100 + \sqrt{x} \right )^{2}$ ?

Answer

Use the square of a binomial pattern as follows:

$\left (100 + \sqrt{x} \right )^{2}$

$= 100^{2}+ 2 \cdot 100 \cdot \sqrt {x} +(\sqrt{x})^{2}$

$= 10,000+ 2 00 \sqrt {x} +x$

This expression is not equivalent to any of the choices.

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Question

$y = (x+ 3)^{2}$

Express in terms of .

Answer

$y = (x+ 3)^{2}$

$= x^{2}+ 2 \cdot 3 \cdot x + 3^{2}$

$= x^{2}+ 6 x + 9$

, so

$16z = x^{2}+ 6 x + 9$

$\frac{16z}{16} = \frac{x^{2}+ 6 x + 9}{16}$

$z = \frac{1}{16}x^{2}+ \frac{3}{8}x+ \frac{9}{16}$

, so

$= \frac{1}{16}x^{2}+ \frac{3}{8}x+ \frac{9}{16}+6$

$= \frac{1}{16}x^{2}+ \frac{3}{8}x+ \frac{9}{16} + \frac{96}{16}$

$= \frac{1}{16}x^{2}+ \frac{3}{8}x+ \frac{105}{16}$

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Question

$a ^{3} = -17$ . Which is the greater quantity?

(a) $a^{6}$

(b)

Answer

By the Power of a Power Principle,

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

Therefore,

$a ^{6} = (a ^{3} )^{2} = (-17)^{2} = 17^{2} = 17 \cdot 17 = 289$

It follows that $a ^{6} > - 34$

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Question

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

(a) $t^{3}$

(b) 11

Answer

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

or

.

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

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Question

$x ^{4} = 12$

$y^{2} = 8$

$(xy) ^{4} = ?$

Answer

By the Power of a Product Principle,

$(xy) ^{4} = x^{4} \cdot y ^{4}$

Also, by the Power of a Power Principle

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

Therefore,

$(xy) ^{4} = x^{4} \cdot (y^{2})^{2} = 12 \cdot 8 ^{2} = 12 \cdot 64 = 768$

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Question

is a negative number. Which is the greater quantity?

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

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

Answer

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

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Question

$t ^{3} = -24$

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

Answer

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,

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

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Question

and are both real numbers.

$t^{4} = 121$

$u ^{4} = 81$

Evaluate .

Answer

, 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

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

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Question

Which is the greater quantity?

(a)

(b)

Answer

The absolute value of is 4, so either or . Likewise, or . However, since $(-4)^{2} = 4 ^{2} = 16$ and $(-3)^{2} = 3 ^{2} = 9$ , it follows that regardless, $t ^{2} = 16$ and $u^{2} = 9$ .

As the product of the sum and the difference of the same two expressions, can be rewritten as the difference of the squares of the expressions:

$= (4t)^{2}-(3u)^{2}$

Using the Power of a Product Principle:

$(4t)^{2}-(3u)^{2}$

$= 4^{2}t^{2}-3^{2}u^{2}$

$=16t^{2}-9u^{2}$

Substituting,

$16t^{2}-9u^{2}$

$=16 \cdot 16 -9 \cdot 9$

Similarly,

$= (3t)^{2}-(4u)^{2}$

$=3 ^{2}t^{2}-4^{2}u^{2}$

$=9t^{2}-16u^{2}$

$=9 \cdot 16 -16 \cdot 9$

Therefore, .

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Question

$t ^{6}= 81$

$u ^{6}= 36$

Which is the greater quantity?

(a) $(t+ u) (t^{2}-tu+u^{2})$

(b) 16

Answer

Multiply the polynomials through distribution:

$(t+ u) (t^{2}-tu+u^{2})$

$= t (t^{2}-tu+u^{2}) + u (t^{2}- tu+u^{2})$

$= t \cdot t^{2}- t \cdot tu+ t \cdot u^{2} + u \cdot t^{2}- u \cdot tu + u \cdot u^{2}$

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

Collecting like terms, the above becomes

$t^{3}+ t^{2} u- t^{2} u + t u^{2} - tu^{2} + u^{3}$

$= t^{3} + u^{3}$

By the Power of a Power Principle,

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

This makes $t ^{3}$ a square root (positive or negative) of $t ^{6}$ , or 81, so

$t^{3} = - \sqrt{81} = -9$

or

$t^{3} = \sqrt{81} = 9$

We can not eliminate either since an odd power of a number can have any sign, and we are not given the sign of .

By similar reasoning, either

$u^{3} = - \sqrt{36} = -6$

or

$u^{3} = \sqrt{36} = 6$

$(t+ u) (t^{2}-tu+u^{2})$ can assume one of four values, depending on which values of $t^{3}$ and $u^{3}$ are selected:

$(t+ u) (t^{2}-tu+u^{2}) = t^{3} + u^{3} = 9 + 6 = 15$

$(t+ u) (t^{2}-tu+u^{2}) = t^{3} + u^{3} = 9 +( -6) = 3$

$(t+ u) (t^{2}-tu+u^{2}) = t^{3} + u^{3} = - 9 +6= -3$

$(t+ u) (t^{2}-tu+u^{2}) = t^{3} + u^{3} = - 9 +( -6) = -15$

Regardless of the choice of $t^{3}$ and $u^{3}$ , $16> (t+ u) (t^{2}-tu+u^{2})$ .

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Question

Which is the greater quantity?

(a) $(t- u) (t^{2}+ tu+u^{2})$

(b) 37

Answer

Multiply the polynomials through distribution:

$(t- u) (t^{2}+ tu+u^{2})$

$= t (t^{2}+ tu+u^{2}) - u (t^{2}+ tu+u^{2})$

$= t \cdot t^{2}+ t \cdot tu+ t \cdot u^{2} - u \cdot t^{2}- u \cdot tu - u \cdot u^{2}$

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

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

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

The absolute value of is 4, so either or . Likewise, or .

If and , we see that

$(t- u) (t^{2}+ tu+u^{2})$

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

$=(-4 )^{3} - 3^{3}$

If and , we see that

$(t- u) (t^{2}+ tu+u^{2})$

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

$=4^{3} - (- 3)^{3}$

In the first scenario, $(t- u) (t^{2}+ tu+u^{2}) < 37$ ; in the second, $(t- u) (t^{2}+ tu+u^{2}) > 37$ . This makes the information insufficient.

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Question

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)

Answer

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

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Question

Simplify:

$(\frac{2x^5}{3y^2})^{-4}$

Answer

First, recognize that raising the fraction to a negative power is the same as raising the inverted fraction to a positive power.

$(\frac{2x^5}{3y^2})^{-4}=(\frac{3y^2}{2x^5})^4$

Apply the exponent within the parentheses and simplify.

$\frac{(3y^2)^4}{(2x^5)^4}$

$\frac{3^4(y^2)^4}{2^4(x^5)^4}$

$\frac{81y^{(2\times 4)}}{16x^{(5\times 4)}}$

$\frac{81y^8}{16x^{20}}$

This fraction cannot be simplified further.

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