Number - Laws of Indices including Zero and Negative Exponents

Simplify the expression: $x^7 \times x^{-3} \div x^2$

$x^{7 + (-3) - 2} = x^2$

Using the multiplication law, we add the indices $7$ and $-3$. Then, using the division law, we subtract $2$ from the result.

Evaluate the numerical value of $5^{-2} \times 25^0$

$\frac{1}{5^2} \times 1 = \frac{1}{25}$

First, convert the negative exponent using $a^{-n} = \frac{1}{a^n}$ to get $\frac{1}{25}$. Then, apply the zero index law where $25^0 = 1$. Finally, $\frac{1}{25} \times 1 = \frac{1}{25}$.

Simplify $\frac{(2a^3)^4}{4a^{10}}$

$\frac{2^4 \times (a^3)^4}{4a^{10}} = \frac{16a^{12}}{4a^{10}} = 4a^2$

First, apply the power to both terms in the bracket: $2^4 = 16$ and $(a^3)^4 = a^{12}$. Then divide the coefficients $16 \div 4 = 4$ and subtract the indices for the base $a$: $12 - 10 = 2$.

Find the value of $n$ if $\left(\frac{1}{3}\right)^n = 27$

$n = -3$

Express $27$ as a power of $3$: $27 = 3^3$. Since $\frac{1}{3} = 3^{-1}$, the equation becomes $(3^{-1})^n = 3^3$. This simplifies to $3^{-n} = 3^3$. Equating the powers, $-n = 3$, so $n = -3$.