How to subtract exponents

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PSAT Math › How to subtract exponents

Questions 1 - 4

1

If m and n are integers such that m < n < 0 and _m_2 – _n_2 = 7, which of the following can be the value of m + n?

I. –5

II. –7

III. –9

I only

II only

I and II only

II and III only

I, II and III only

Explanation

m and n are both less than zero and thus negative integers, giving us _m_2 and _n_2 as perfect squares. The only perfect squares with a difference of 7 is 16 – 9, therefore m = –4 and n = –3.

2

$\frac{7^5+7^4}{8}=$

$\frac{7^9}{8}$

$\frac{7}{8}$

$\frac{14^4}{8}$

$\frac{7^4}{8}$

Explanation

To simplify, we can rewrite the numerator using a common exponential base.

$\frac{7^5+7^4}{8}=\frac{7(7^4)+7^4}{8}$

Now, we can factor out the numerator.

$\frac{7(7^4)+7^4}{8}=\frac{7^4(7+1)}{8}=\frac{7^4(8)}{8}$

The eights cancel to give us our final answer.

$\frac{7^4(8)}{8}=7^4$

3

Solve:

$4^{8}\div2^{10}=$

Explanation

$4^{8}=\left ( 2^{2}\right )^{8}=2^{16}$

Subtract the denominator exponent from the numerator's exponent, since they have the same base.

$\frac{2^{16}}{2^{10}}=2^{16-10}=2^{6}=64$

4

If $x\star y=x^2-y^2$ , then what is $4\star (5\star 3)$ ?

Explanation

$4\star (5\star 3)$

Follow the order of operations by solving the expression within the parentheses first.

$x\star y=x^2-y^2$

$5\star 3=(5)^2-(3)^2$

$5\star 3=25-9=16$

Return to solve the original expression.

$4\star(5\star3)=4\star(16)$

$x\star y=x^2-y^2$

$4\star16=(4)^2-(16)^2$

$4\star(16)=-240$

$4\star(5\star3)=-240$

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