A Negative Power Is Only an Instruction to Turn the Fraction Over

It means take the reciprocal — nothing more alarming than that.

$$2^{-3}=\frac{1}{2^3}=\frac{1}{8}$$

A negative sign in the exponent has nothing to do with the answer being negative. $2^{-3}$ is a positive number; it is simply a small one. And for a fraction the effect is even simpler:

$${\left(\frac{2}{3}\right)}^{-4}={\left(\frac{3}{2}\right)}^4$$

The negative exponent just turns the fraction over and becomes positive. This page covers the first part of the ICSE Class 8 Mathematics chapter on exponents: what the three kinds of exponent mean, the laws that combine them, and how to clear a negative exponent away.

In $a^m$, the number $a$ is the base and $m$ is the exponent, index or power.

A positive exponent tells you how many times to use the base as a factor:

$$5^4=5\times 5\times 5\times 5=625$$

This is the expanded form, and converting between the two directions is the first skill of the chapter:

$$2\times 2\times 2\times 2\times 2=2^5=32$$

A zero exponent always gives $1$:

$$a^0=1(a\ne 0)$$

So $7^0=1$, $(-19{)}^0=1$ and ${\left(\frac{3}{8}\right)}^0=1$. Why it must be $1$ follows from the division law: $\frac{a^3}{a^3}$ is clearly $1$, and by the law it is also $a^{3-3}=a^0$. The two answers must agree.

A negative exponent means the reciprocal of the positive power:

$$a^{-m}=\frac{1}{a^m},\text{and equivalently}\frac{1}{a^{-m}}=a^m$$

$$3^{-2}=\frac{1}{3^2}=\frac{1}{9},10^{-3}=\frac{1}{1000}=0.001$$

For a rational base the rule becomes the flip:

$${\left(\frac{a}{b}\right)}^{-m}={\left(\frac{b}{a}\right)}^m$$

Expanded form using powers of ten. Any number can be written as a sum of digits multiplied by powers of $10$:

$$4307=4\times 10^3+3\times 10^2+0\times 10^1+7\times 10^0$$

Checking: $4000+300+0+7=4307$. Correct — and notice that the units digit needs $10^0=1$, which is where the zero exponent earns its keep.

A sign distinction that costs marks. $(-2{)}^4$ and $-2^4$ are different. In the first, the bracket means the whole of $-2$ is the base, so $(-2{)}^4=16$, positive. In the second, only the $2$ is raised, and the minus applies afterwards, so $-2^4=-16$. A negative base raised to an even power gives a positive answer and to an odd power a negative one.

Use the laws to collect everything into a single power, then apply $a^{-m}=\frac{1}{a^m}$ once at the end.

Worked example 1 — same base throughout. Simplify $\frac{3^{-2}\times 3^5}{3^4}$.

Add the exponents on top, then subtract the one below:

$$3^{-2+5-4}=3^{-1}=\frac{1}{3}$$

One line of exponent arithmetic replaces all the multiplying and dividing.

Worked example 2 — a rational base. Evaluate ${\left(\frac{2}{3}\right)}^{-4}$.

Flip the fraction and make the exponent positive:

$${\left(\frac{3}{2}\right)}^4=\frac{3^4}{2^4}=\frac{81}{16}$$

Worked example 3 — a difference of two reciprocals. Evaluate ${\left(\frac{1}{2}\right)}^{-3}-{\left(\frac{1}{3}\right)}^{-2}$.

Flip each one:

$${\left(\frac{2}{1}\right)}^3-{\left(\frac{3}{1}\right)}^2=2^3-3^2=8-9=-1$$

The answer is negative, which is worth noticing: negative exponents never make a result negative, but ordinary subtraction can.

Worked example 4 — a negative rational base. Evaluate ${\left(-\frac{2}{3}\right)}^{-2}\times {\left(\frac{3}{2}\right)}^{-3}$.

Flip each factor:

$${\left(-\frac{3}{2}\right)}^2\times {\left(\frac{2}{3}\right)}^3=\frac{9}{4}\times \frac{8}{27}$$

The first is positive because the exponent $2$ is even. Now multiply:

$$\frac{9\times 8}{4\times 27}=\frac{72}{108}=\frac{2}{3}$$

Dividing numerator and denominator by the HCF $36$ gives $\frac{2}{3}$.

Worked example 5 — collecting different bases. Simplify $\frac{2^{-3}\times 3^2}{2^{-5}}$.

Deal with each base separately. For the twos, $2^{-3}÷2^{-5}=2^{-3-(-5)}=2^2$. So

$$2^2\times 3^2=4\times 9=36$$

The order of work that avoids mistakes. Convert negative exponents to positive last, not first. Doing the exponent arithmetic while the powers are still in index form keeps the numbers small — in example 1, working with $3^{-2}$, $3^5$ and $3^4$ as fractions first would have meant handling $\frac{1}{9}$, $243$ and $81$ for no reason.

And a check that costs nothing. A negative exponent on a base greater than $1$ always gives an answer **less than $1$, and a negative exponent on a proper fraction always gives an answer greater than $1$**. If ${\left(\frac{2}{3}\right)}^{-4}$ had come out smaller than $1$, something would have gone wrong.