We've already looked at the power of a power rule, which states:
$\left(a^m\right)^n=a^{mn}$(am)n=amn
Now we are going to use apply this rule to terms involving negative bases, and also negative exponents.
The base number stays the same. We just multiply the powers.
When we raise a negative numbers by an even power, we get a positive answer e.g. $\left(-4\right)^2=-4\times\left(-4\right)$(−4)2=−4×(−4)$=$=$16$16
The same pattern applies to negative powers: $\left(-2\right)^{-4}=2^{-4}$(−2)−4=2−4$=$=$\frac{1}{16}$116
This means we can simplify expressions involving negative numbers with negative exponents, and express them as a fraction with a positive exponent. If the power is even the fraction will be positive, but if the power is odd the fraction will be negative.
For example, $\left(-5\right)^{-3}=\frac{1}{\left(-5\right)^3}$(−5)−3=1(−5)3 $=$= $-\frac{1}{125}$−1125
Simplify the following into the form $a^b$ab:
$\left(6^7\right)^{-3}$(67)−3
Evaluate $\left(5^{-9}\right)^0$(5−9)0.
$\left(\left(-4\right)^{-8}\right)^2$((−4)−8)2 simplifies to which of the following:
$-4^{-6}$−4−6
$4^{-6}$4−6
$4^{-16}$4−16
$-4^{-16}$−4−16
We want to simplify the following expression using exponent laws: $4^{-2}\times64^{-3}$4−2×64−3.
To use the exponent laws, we need both terms to be written with the same base.
Fill in the box to re-write $64^{-3}$64−3 with a base of $4$4.
$64^{-3}=\left(4^{\editable{}}\right)^{-3}$64−3=(4)−3
Using the result of the previous part, express $4^{-2}\times64^{-3}$4−2×64−3 in simplest positive exponential form.