# Exercises on exponents and powers

We offer free exercises on exponents and powers of numbers. You can take a look at the chapter exponents first.

We recall that the powers of a number $a$ are defined by $a^n$ with $n$ integer. This is, $a^n=a\times a\times\cdots\times a$ (the multiplication of $a$ n times).

Exercise: Write $1800$ as product of powers:

Solution: Notice that $1800$ is an even number and $1800=200\times 9= 8 \times 25\times 3^2=2^3\times 3^2\times 5^2$. Thus
\begin{align*}
1800=2^3\times 3^2\times 5^2.
\end{align*}

exercise: Compute the number
\begin{align*}
\left(\frac{2}{3^2}\right)^{-3}\times \frac{2^2}{3^4}.
\end{align*}
Solution: Notice that
\begin{align*}
\left(\frac{2}{3^2}\right)^{-3}=\left(\frac{3^2}{2}\right)^{3}=\frac{(3^2)^3}{2^3}=\frac{3^6}{2^3}.
\end{align*}
Then
\begin{align*}
\left(\frac{2}{3^2}\right)^{-3}\times \frac{2^2}{3^4}=\frac{3^6}{2^3}\times \frac{2^2}{3^4}=\frac{3^2}{2}=\frac{9}{2}.
\end{align*}

Exercise: Let $a,b$ and $c$ be non zero numbers.   Simplify the number
\begin{align*}
A=\frac{ab^{-2}(bc^{-1})^{-2}b^3 c^2}{a^{-3}b^5 \left(\frac{1}{a^2}c^3\right)^2}.
\end{align*}Compute $A$ in the case: $a=10^3,\;b=10^2$ and $c=0.0001$.

Preuve
Exercise 1: Simplify the number
\begin{align*}
A=1+\frac{1}{2}+\frac{1}{4}+\cdots+\frac{1}{32}.
\end{align*}
Solution: Observe that $32=2\times 4\times 4=2^5$. Then $A$ can be rewritten as
\begin{align*}
A&=1+\frac{1}{2}+\frac{1}{2^2}+\cdots+\frac{1}{2^5}\cr &=1+\frac{1}{2}+\left(\frac{1}{2}\right)^2+\cdots+\left(\frac{1}{2}\right)^5.
\end{align*}
Now we will use a remarkable identity of order 6. Then
\begin{align*}
1-\left(\frac{1}{2}\right)^6=\left(1-\frac{1}{2}\right)\left(1+\frac{1}{2}+\left(\frac{1}{2}\right)^2+\cdots+\left(\frac{1}{2}\right)^5\right)=\frac{1}{2} A
\end{align*}
On the other hand, we compute
\begin{align*}
1-\left(\frac{1}{2}\right)^6=1-\frac{1}{64}=\frac{63}{64}.
\end{align*}
Thus
\begin{align*}
\frac{1}{2} A=\frac{63}{64}.
\end{align*}
That is $A=\frac{63}{32}$.

Exercise 2: Let $a$ and $b$ be two distinct numbers, and $n$ a relative number. Simplify the expression
\begin{align*}
B=\frac{a^n b-a^{n+1}}{ab^n-b^{n+1}}\times \left(\frac{a}{b}\right)^{-n}.
\end{align*}
Solution: We factor the numerator by $a^n$ and the denominator by $b^n$ in the fraction we obtain
\begin{align*}
B&=\frac{a^n( b-a)}{b^n(a-b)}\times \left(\frac{a}{b}\right)^{-n}
\cr &=\left(\frac{a}{b}\right)^{-n}\frac{( b-a)}{(a-b)}\times \left(\frac{a}{b}\right)^{-n}\cr &= - \left(\frac{a}{b}\right)^{n-n}=-1 \times 1=-1.
\end{align*}