Chapter 18, Problem 1P

Estimate the common logarithm of 10 using linear interpolation.

(a) Interpolate between $\log 8=0.9030900\text{ and log }12=1.0791812$.

(b) Interpolate between $\log 9=0.9542425\text{ and log }11=1.0413927$.

For each of the interpolations, compute the percent relative error based on the true value.

To determine

To calculate: The common logarithmic of $10$ by the linear interpolation between $\log 8=0.9030900$ and $\log 12=1.0791812$. Also, find the percentage relative error based on the true value.

Answer to Problem 1P

Solution:

The common logarithmic of $10$ by the linear interpolation between $\log 8$ and $\log 12$ is $0.9911356$ with relative percentage error of $0.88644\%$.

Explanation of Solution

Given Information:

The values,

$\begin{array}{c}\log 8=0.9030900\\ \log 12=1.0791812\end{array}$

Formula used:

Linear interpolation formula:

$f_1\left(x\right)=f\left(x_0\right)+\frac{f\left(x_1\right)-f\left(x_0\right)}{x_1-x_0}\left(x-x_0\right)$

And, formula for percentage relative error is,

${\epsilon }_t=\left|\frac{\text{True value}-\text{Approximate}\text{value}}{\text{True value}}\right|\times 100\%$

Calculation:

Consider the values, $\log 8=0.9030900$ and $\log 12=1.0791812$.

Here, $x_0=8$ and $x_1=12$. Therefore,

$\begin{array}{l}f\left(x_0\right)=0.9030900\\ f\left(x_1\right)=1.0791812\end{array}$

Thus, the value of log 10 by the linear interpolation is,

$\begin{array}{c}{f}_1\left(10\right)=0.9030900+\frac{1.0791812-0.9030900}{12-8}\left(10-8\right)\\ =0.9030900+\frac{0.1760912}{4}\times 2\\ =0.9030900+0.0880456\\ =0.9911356\end{array}$

Now, the true value of $\log 10=1$. Therefore, relative percentage error is,

$\begin{array}{c}{\epsilon }_t=\frac{1-0.9911356}{1}\times 100\%\\ =0.0088644\times 100\%\\ =0.88644\%\end{array}$

Hence, the value of log 10 by the linear interpolation is $0.9911356$ with relative percentage error of $0.88644\%$.

To determine

To calculate: The common logarithmic of $10$ by the linear interpolation between $\log 9=0.9542425$ and $\log 11=1.0413927$. Also, find the percentage relative error based on the true value.

Answer to Problem 1P

Solution:

The common logarithmic of $10$ by the linear interpolation between $\log 9$ and $\log 11$ is $0.9978176$ with relative percentage error of $0.218\%$.

Explanation of Solution

Given Information:

The values,

$\begin{array}{c}\log 9=0.9542425\\ \log 11=1.0413927\end{array}$

Formula used:

Linear interpolation formula:

$f_1\left(x\right)=f\left(x_0\right)+\frac{f\left(x_1\right)-f\left(x_0\right)}{x_1-x_0}\left(x-x_0\right)$

And, formula for percentage relative error is,

${\epsilon }_t=\left|\frac{\text{True value}-\text{Approximate}\text{value}}{\text{True value}}\right|\times 100\%$

Calculation:

Consider the values, $\log 9=0.9542425$ and $\log 11=1.0413927$.

Here, $x_0=9$ and $x_1=11$. Therefore,

$\begin{array}{l}f\left(x_0\right)=0.9542425\\ f\left(x_1\right)=1.0413927\end{array}$

Thus, the value of log 10 by the linear interpolation is,

$\begin{array}{c}{f}_1\left(10\right)=0.9542425+\frac{1.0413927-0.9542425}{11-9}\left(10-9\right)\\ =0.9542425+\frac{0.0871502}{2}\times 1\\ =0.9542425+0.0435751\\ =0.9978176\end{array}$

Now, the true value of $\log 10=1$. Therefore, relative percentage error is,

$\begin{array}{c}{\epsilon }_t=\left|\frac{1-0.9978176}{1}\right|\times 100\%\\ =0.0021824\times 100\%\\ =0.218\%\end{array}$

Hence, the value of log 10 by the linear interpolation is $0.9978176$ with relative percentage error of $0.218\%$.