How to add exponents

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ISEE Upper Level Quantitative Reasoning › How to add exponents

Questions 1 - 10

What is the value of the expression:

$x^{3}\cdotx^{y}\cdot x^{2+y}$

$x^{4}\cdotx^{y+2}$

$x^{4}\cdotx^{y+4}$

$x^{6}\cdotx^{y+2}$

$x^{3}\cdotx^{y+2}$

Explanation

When values, having the same base, are multiplied by one another, the exponents are added together and the base stays the same.

Thus,

$x^{3}\cdotx^{y}\cdot x^{2+y}$

is equal to

$x^{4}\cdotx^{y+2}$

Simplify the expression:

$3x^{4}+6x^{3}-9x^{2}$

$3x^{2}(x^{2}+2x-3)$

$3x^{2}(x^{2}+2x-3x)$

$2x^{2}(x^{2}+2x-3x)$

$3x(x^{2}+2x-3)$

$3x(x^{2}+3x-3)$

Explanation

To simplify this problem we need to factor out a

$3x^{4}+6x^{3}-9x^{2}$

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

We can do this because multiplying exponents is the same as adding them. Therefore,

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

$=3x^2 \cdot x^2 +3x^2\cdot2x -3x^2\cdot3=3x^{2+2}+3\cdot2x^{2+1}-3\cdot 3x^2$

Simplify:

$2x^{2}\cdot 3x^{4}$

$6x^{6}$

$6x^{8}$

$5x^{6}$

$6x^{2}$

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

Explanation

When multiplying exponents, the exponents are added together.

$2x^{2}\cdot 3x^{4}=6x^{2+4}=6x^{6}$

Simplify:

$(6a^3)(\frac{1}{3}a^2)$

Explanation

Based on the product rule for exponents in order to multiply two exponential terms with the same base, add their exponents:

$(x)^a(x)^b=x^{a+b}$

So we can write:

$(6a^3)\left(\frac{1}{3}a^2\right)=\left(\frac{6}{3}\right)\times a^{3+2}=2a^5$

Solve: $3^5 \cdot 3^{-8}$

$\frac{1}{27}$

$\frac{1}{3^{13}}$

$3^{13}$

$\frac{1}{9}$

Explanation

It is not necessary to evaluate both terms and multiply.

According to the rules of exponents, when we have the same bases raised to some power that are multiplied with each other, we can add the powers.

$3^5 \cdot 3^{-8} = 3^{5+( -8)} = 3^{-3}$

This term can be rewritten as a fraction.

$3^{-3 } = \frac{1}{3^3} = \frac{1}{27}$

The answer is: $\frac{1}{27}$

Simplify: $x^{7}\cdot2x^{3}$

$2x^{10}$

$x^{10}$

$2x^{21}$

$x^{21}$

Explanation

When multiplying exponents, the exponents are added together. Thus, 3 and 7 are added together for a sum of 10. In this problem, the "2" becomes a coefficient in front of the x. Therefore, the correct answer is:

$2x^{10}$

Simplify:

Explanation

In order to add exponential terms, both the base and the exponent must be the same. So we can write:

Which of the following is equivalent to the expression below?

$4^{x}\cdot2^{5}$

$2^{2x+5}$

$8^{5x}$

$6^{5x}$

$8^{10x}$

Explanation

When exponents are multiplied by one another, and the base is the same, the exponents can be added together.

The first step is to try to create a common base.

$4^{x}\cdot2^{5}$

Given that the square of 2 is for, the expression can be rewritten as:

$2^{2}^{x}\cdot 2^{5}$

$2^{2x+5}$

Which is the greater quantity?

(A) The sum of the first ten perfect square integers

(B) The sum of the first five perfect cube integers

(A) is greater

(B) is greater

(A) and (B) are equal

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

Explanation

The sum of the first ten perfect square integers:

$\begin{matrix} ;; ; 1\ ;; ; 4\ ;; ; 9\ ;16\ ;25\ ;36\ ;49\ ;64\ ;81\ \underline{100}\ 385 \end{matrix}$

The sum of the first five perfect cube integers:

$\begin{matrix} ;; ; 1\ ;; ; 8\ ;27\ ;64\ \underline{125}\ 225 \end{matrix}$

(A) is greater.

Two quantities are given - one in Column A and the other in Column B. Compare the quantities in the two columns.

Assume, in both columns, that .

Column A Column B

The quantity in Column B is greater.

The quantity in Column A is greater.

The quantities in both columns are equal.

The relationship cannot be determined from the info given.

Explanation

When you are adding and subtracting terms with exponents, you combine like terms. Since both columns have expressions with the same exponent throughout, you are good to just look at the coefficients. Remember, a coefficient is the number in front of a variable. Therefore, Column A is since . Column B is since . We can see that Column B is greater.

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