Laws of Exponents: Practice

Same base throughout, so add all of the exponents at once.

$2^3 \times 2^4 \times 2^2 = 2^{3+4+2} = 2^9$

Combine the top with the product rule, then divide with the quotient rule.

$\frac{5^7 \times 5^2}{5^4} = \frac{5^9}{5^4} = 5^{9-4} = 5^5$

Since $6 = 2 \times 3$, use the power-of-a-product rule.

$6^5 = (2 \times 3)^5 = 2^5 \times 3^5$

The exponent lands on each factor separately.

Resolve the power of a power on top, then divide.

$(4^3)^2 = 4^6, \qquad \frac{4^6}{4^4} = 4^{6-4} = 4^2$

Multiply inside the parentheses first, then cube. (The power-of-a-product rule gives the same answer: $3^3 \times 2^3 = 27 \times 8 = 216$.)

$(3 \times 2)^3 = 6^3 = 216$

The product rule adds the exponents, so the missing exponent plus $3$ must equal $9$.

$3 + \square = 9 \;\Rightarrow\; \square = 6$

Check: $2^3 \times 2^6 = 2^9$.

A power of a quotient raises the numerator and denominator separately.

$\left(\frac{2}{3}\right)^3 = \frac{2^3}{3^3} = \frac{8}{27}$

Add the exponents on top, then subtract the bottom exponent.

$\frac{10^4 \times 10^3}{10^5} = \frac{10^7}{10^5} = 10^{7-5} = 10^2$

A power of a product gives the exponent to each factor.

$(2 \times 5)^4 = 2^4 \times 5^4$

The power rule multiplies the exponents, so the inner base raised to the $2$ must give exponent $6$ after multiplying by the outer $3$.

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

So the blank is $a$, since a plain $a$ is $a^1$ and $(a^2)^3 = a^6$.

Combine the powers of $3$ with the product rule. The $5^3$ has a different base, so the product rule cannot absorb it.

$3^4 \times 3^2 = 3^6 \;\Rightarrow\; 3^4 \times 3^2 \times 5^3 = 3^6 \times 5^3$

The product rule needs a shared base, so the $5^3$ stays where it is.

Resolve the power of a power, then add the exponent of the lone factor ($7 = 7^1$).

$(7^2)^2 = 7^4, \qquad 7^4 \times 7^1 = 7^{4+1} = 7^5$