Chapter 4.5, Problem 89E

To determine

Find two whole numbers between which the solution must lie.

Answer to Problem 89E

  $\left[f\left(1\right)<0,f\left(2\right)>0\right]$

  $\boxed{x\approx 1.3636}$

Explanation of Solution

Given information:

Estimating a Solution without actually solving the equation, find two whole numbers between which the solution of $9^x=20$ must lie. Do the same for $9^x=100$ . Explain how you reached your conclusions

Calculation:

Consider the given equation,

  $9^x=20$

Let

  $f\left(x\right)=9^x-20$

Root of $f\left(x\right)=0$ so root of $9^x=20$

Substitute $1$ for $x$ , in $f\left(x\right)$

  $\begin{array}{l}f\left(1\right)=9^1-20\\ f\left(1\right)=-11<0\end{array}$

Substitute $2$ for $x$ , in $f\left(x\right)$

  $\begin{array}{l}f\left(2\right)=9^2-20\\ f\left(2\right)=81-20\\ f\left(2\right)=61>0\end{array}$

So, by intermediate value theorem the roots must lie between number $1$ and $2$

Since, $\left[f\left(1\right)<0,f\left(2\right)>0\right]$

Let,

  $g\left(x\right)=9^x-100$

Root of $g\left(x\right)=0$ so root of $9^x=100$

Substitute $1$ for $x$ , in $g\left(x\right)$

  $\begin{array}{l}g\left(1\right)=9^1-100\\ =9-100\\ =-91<0\end{array}$

Substitute $2$ for $x$ , in $g\left(x\right)$

  $\begin{array}{l}g\left(2\right)=9^2-100\\ =81-100\\ =-19<0\end{array}$

Substitute $3$ for $x$ , in $g\left(x\right)$

  $\begin{array}{l}g\left(3\right)=9^3-100\\ =729-100\\ =629>0\end{array}$

So, by intermediate value theorem the roots must lie between number $2$ and $3$

Since, $\left[g\left(2\right)<0,g\left(3\right)>0\right]$

Now, to check the answer for first case. take log on both sides,

  $\begin{array}{l}{9}^x=20\\ \log 9^x=\log 20\\ x\log 9=\log 20\\ x=\frac{\log 20}{\log 9}\\ x\approx 1.3634\end{array}$

Hence, intermediate value theorem for first case is correct.

Hence, the solution is, $\left[f\left(1\right)<0,f\left(2\right)>0\right]$

  $\boxed{x\approx 1.3636}$