Estimate the common logarithm of 10 using linear interpolation. (a) Interpolate between log 8 = 0.9030900 and log 12 = 1.0791812 . (b) Interpolate between log 9 = 0.9542425 and log 11 = 1.0413927 . For each of the interpolations, compute the percent relative error based on the true value.

Question

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Answer 1

Textbook Question

Chapter 18, Problem 1P

Estimate the common logarithm of 10 using linear interpolation.

(a) Interpolate between $log 8 = 0.9030900 and log 12 = 1.0791812$ .

(b) Interpolate between $log 9 = 0.9542425 and log 11 = 1.0413927$ .

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

(a)

Expert Solution

To determine

To calculate: The common logarithmic of $10$ by the linear interpolation between $log8=0.9030900$ and $log12=1.0791812$ . 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 $log8$ and $log12$ is $0.9911356$ with relative percentage error of $0.88644%$ .

Explanation of Solution

Given Information:

The values,

$log8=0.9030900log12=1.0791812$

Formula used:

Linear interpolation formula:

$f1(x)=f(x0)+f(x1)−f(x0)x1−x0(x−x0)$

And, formula for percentage relative error is,

$εt=|True value−Approximate valueTrue value|×100%$

Calculation:

Consider the values, $log8=0.9030900$ and $log12=1.0791812$ .

Here, $x0=8$ and $x1=12$ . Therefore,

$f(x0)=0.9030900f(x1)=1.0791812$

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

$f1(10)=0.9030900+1.0791812−0.903090012−8(10−8)=0.9030900+0.17609124×2=0.9030900+0.0880456=0.9911356$

Now, the true value of $log10=1$ . Therefore, relative percentage error is,

$εt=1−0.99113561×100%=0.0088644×100%=0.88644%$

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

(b)

Expert Solution

To determine

To calculate: The common logarithmic of $10$ by the linear interpolation between $log9=0.9542425$ and $log11=1.0413927$ . 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 $log9$ and $log11$ is $0.9978176$ with relative percentage error of $0.218%$ .

Explanation of Solution

Given Information:

The values,

$log9=0.9542425log11=1.0413927$

Formula used:

Linear interpolation formula:

$f1(x)=f(x0)+f(x1)−f(x0)x1−x0(x−x0)$

And, formula for percentage relative error is,

$εt=|True value−Approximate valueTrue value|×100%$

Calculation:

Consider the values, $log9=0.9542425$ and $log11=1.0413927$ .

Here, $x0=9$ and $x1=11$ . Therefore,

$f(x0)=0.9542425f(x1)=1.0413927$

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

$f1(10)=0.9542425+1.0413927−0.954242511−9(10−9)=0.9542425+0.08715022×1=0.9542425+0.0435751=0.9978176$

Now, the true value of $log10=1$ . Therefore, relative percentage error is,

$εt=|1−0.99781761|×100%=0.0021824×100%=0.218%$

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

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Answer 2

Textbook Question

Chapter 18, Problem 1P

Estimate the common logarithm of 10 using linear interpolation.

(a) Interpolate between $log 8 = 0.9030900 and log 12 = 1.0791812$ .

(b) Interpolate between $log 9 = 0.9542425 and log 11 = 1.0413927$ .

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

(a)

Expert Solution

To determine

To calculate: The common logarithmic of $10$ by the linear interpolation between $log8=0.9030900$ and $log12=1.0791812$ . 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 $log8$ and $log12$ is $0.9911356$ with relative percentage error of $0.88644%$ .

Explanation of Solution

Given Information:

The values,

$log8=0.9030900log12=1.0791812$

Formula used:

Linear interpolation formula:

$f1(x)=f(x0)+f(x1)−f(x0)x1−x0(x−x0)$

And, formula for percentage relative error is,

$εt=|True value−Approximate valueTrue value|×100%$

Calculation:

Consider the values, $log8=0.9030900$ and $log12=1.0791812$ .

Here, $x0=8$ and $x1=12$ . Therefore,

$f(x0)=0.9030900f(x1)=1.0791812$

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

$f1(10)=0.9030900+1.0791812−0.903090012−8(10−8)=0.9030900+0.17609124×2=0.9030900+0.0880456=0.9911356$

Now, the true value of $log10=1$ . Therefore, relative percentage error is,

$εt=1−0.99113561×100%=0.0088644×100%=0.88644%$

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

(b)

Expert Solution

To determine

To calculate: The common logarithmic of $10$ by the linear interpolation between $log9=0.9542425$ and $log11=1.0413927$ . 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 $log9$ and $log11$ is $0.9978176$ with relative percentage error of $0.218%$ .

Explanation of Solution

Given Information:

The values,

$log9=0.9542425log11=1.0413927$

Formula used:

Linear interpolation formula:

$f1(x)=f(x0)+f(x1)−f(x0)x1−x0(x−x0)$

And, formula for percentage relative error is,

$εt=|True value−Approximate valueTrue value|×100%$

Calculation:

Consider the values, $log9=0.9542425$ and $log11=1.0413927$ .

Here, $x0=9$ and $x1=11$ . Therefore,

$f(x0)=0.9542425f(x1)=1.0413927$

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

$f1(10)=0.9542425+1.0413927−0.954242511−9(10−9)=0.9542425+0.08715022×1=0.9542425+0.0435751=0.9978176$

Now, the true value of $log10=1$ . Therefore, relative percentage error is,

$εt=|1−0.99781761|×100%=0.0021824×100%=0.218%$

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

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Answer 3

Textbook Question

Chapter 18, Problem 1P

Estimate the common logarithm of 10 using linear interpolation.

(a) Interpolate between $log 8 = 0.9030900 and log 12 = 1.0791812$ .

(b) Interpolate between $log 9 = 0.9542425 and log 11 = 1.0413927$ .

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

(a)

Expert Solution

To determine

To calculate: The common logarithmic of $10$ by the linear interpolation between $log8=0.9030900$ and $log12=1.0791812$ . 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 $log8$ and $log12$ is $0.9911356$ with relative percentage error of $0.88644%$ .

Explanation of Solution

Given Information:

The values,

$log8=0.9030900log12=1.0791812$

Formula used:

Linear interpolation formula:

$f1(x)=f(x0)+f(x1)−f(x0)x1−x0(x−x0)$

And, formula for percentage relative error is,

$εt=|True value−Approximate valueTrue value|×100%$

Calculation:

Consider the values, $log8=0.9030900$ and $log12=1.0791812$ .

Here, $x0=8$ and $x1=12$ . Therefore,

$f(x0)=0.9030900f(x1)=1.0791812$

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

$f1(10)=0.9030900+1.0791812−0.903090012−8(10−8)=0.9030900+0.17609124×2=0.9030900+0.0880456=0.9911356$

Now, the true value of $log10=1$ . Therefore, relative percentage error is,

$εt=1−0.99113561×100%=0.0088644×100%=0.88644%$

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

(b)

Expert Solution

To determine

To calculate: The common logarithmic of $10$ by the linear interpolation between $log9=0.9542425$ and $log11=1.0413927$ . 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 $log9$ and $log11$ is $0.9978176$ with relative percentage error of $0.218%$ .

Explanation of Solution

Given Information:

The values,

$log9=0.9542425log11=1.0413927$

Formula used:

Linear interpolation formula:

$f1(x)=f(x0)+f(x1)−f(x0)x1−x0(x−x0)$

And, formula for percentage relative error is,

$εt=|True value−Approximate valueTrue value|×100%$

Calculation:

Consider the values, $log9=0.9542425$ and $log11=1.0413927$ .

Here, $x0=9$ and $x1=11$ . Therefore,

$f(x0)=0.9542425f(x1)=1.0413927$

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

$f1(10)=0.9542425+1.0413927−0.954242511−9(10−9)=0.9542425+0.08715022×1=0.9542425+0.0435751=0.9978176$

Now, the true value of $log10=1$ . Therefore, relative percentage error is,

$εt=|1−0.99781761|×100%=0.0021824×100%=0.218%$

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

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Answer 4

Textbook Question

Chapter 18, Problem 1P

Estimate the common logarithm of 10 using linear interpolation.

(a) Interpolate between $log 8 = 0.9030900 and log 12 = 1.0791812$ .

(b) Interpolate between $log 9 = 0.9542425 and log 11 = 1.0413927$ .

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

(a)

Expert Solution

To determine

To calculate: The common logarithmic of $10$ by the linear interpolation between $log8=0.9030900$ and $log12=1.0791812$ . 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 $log8$ and $log12$ is $0.9911356$ with relative percentage error of $0.88644%$ .

Explanation of Solution

Given Information:

The values,

$log8=0.9030900log12=1.0791812$

Formula used:

Linear interpolation formula:

$f1(x)=f(x0)+f(x1)−f(x0)x1−x0(x−x0)$

And, formula for percentage relative error is,

$εt=|True value−Approximate valueTrue value|×100%$

Calculation:

Consider the values, $log8=0.9030900$ and $log12=1.0791812$ .

Here, $x0=8$ and $x1=12$ . Therefore,

$f(x0)=0.9030900f(x1)=1.0791812$

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

$f1(10)=0.9030900+1.0791812−0.903090012−8(10−8)=0.9030900+0.17609124×2=0.9030900+0.0880456=0.9911356$

Now, the true value of $log10=1$ . Therefore, relative percentage error is,

$εt=1−0.99113561×100%=0.0088644×100%=0.88644%$

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

(b)

Expert Solution

To determine

To calculate: The common logarithmic of $10$ by the linear interpolation between $log9=0.9542425$ and $log11=1.0413927$ . 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 $log9$ and $log11$ is $0.9978176$ with relative percentage error of $0.218%$ .

Explanation of Solution

Given Information:

The values,

$log9=0.9542425log11=1.0413927$

Formula used:

Linear interpolation formula:

$f1(x)=f(x0)+f(x1)−f(x0)x1−x0(x−x0)$

And, formula for percentage relative error is,

$εt=|True value−Approximate valueTrue value|×100%$

Calculation:

Consider the values, $log9=0.9542425$ and $log11=1.0413927$ .

Here, $x0=9$ and $x1=11$ . Therefore,

$f(x0)=0.9542425f(x1)=1.0413927$

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

$f1(10)=0.9542425+1.0413927−0.954242511−9(10−9)=0.9542425+0.08715022×1=0.9542425+0.0435751=0.9978176$

Now, the true value of $log10=1$ . Therefore, relative percentage error is,

$εt=|1−0.99781761|×100%=0.0021824×100%=0.218%$

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

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

(a)

Expert Solution

To determine

To calculate: The common logarithmic of $10$ by the linear interpolation between $log8=0.9030900$ and $log12=1.0791812$ . 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 $log8$ and $log12$ is $0.9911356$ with relative percentage error of $0.88644%$ .

Explanation of Solution

Given Information:

The values,

$log8=0.9030900log12=1.0791812$

Formula used:

Linear interpolation formula:

$f1(x)=f(x0)+f(x1)−f(x0)x1−x0(x−x0)$

And, formula for percentage relative error is,

$εt=|True value−Approximate valueTrue value|×100%$

Calculation:

Consider the values, $log8=0.9030900$ and $log12=1.0791812$ .

Here, $x0=8$ and $x1=12$ . Therefore,

$f(x0)=0.9030900f(x1)=1.0791812$

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

$f1(10)=0.9030900+1.0791812−0.903090012−8(10−8)=0.9030900+0.17609124×2=0.9030900+0.0880456=0.9911356$

Now, the true value of $log10=1$ . Therefore, relative percentage error is,

$εt=1−0.99113561×100%=0.0088644×100%=0.88644%$

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

(b)

Expert Solution

To determine

To calculate: The common logarithmic of $10$ by the linear interpolation between $log9=0.9542425$ and $log11=1.0413927$ . 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 $log9$ and $log11$ is $0.9978176$ with relative percentage error of $0.218%$ .

Explanation of Solution

Given Information:

The values,

$log9=0.9542425log11=1.0413927$

Formula used:

Linear interpolation formula:

$f1(x)=f(x0)+f(x1)−f(x0)x1−x0(x−x0)$

And, formula for percentage relative error is,

$εt=|True value−Approximate valueTrue value|×100%$

Calculation:

Consider the values, $log9=0.9542425$ and $log11=1.0413927$ .

Here, $x0=9$ and $x1=11$ . Therefore,

$f(x0)=0.9542425f(x1)=1.0413927$

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

$f1(10)=0.9542425+1.0413927−0.954242511−9(10−9)=0.9542425+0.08715022×1=0.9542425+0.0435751=0.9978176$

Now, the true value of $log10=1$ . Therefore, relative percentage error is,

$εt=|1−0.99781761|×100%=0.0021824×100%=0.218%$

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

Answer 8

(a)

Expert Solution

To determine

To calculate: The common logarithmic of $10$ by the linear interpolation between $log8=0.9030900$ and $log12=1.0791812$ . 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 $log8$ and $log12$ is $0.9911356$ with relative percentage error of $0.88644%$ .

Explanation of Solution

Given Information:

The values,

$log8=0.9030900log12=1.0791812$

Formula used:

Linear interpolation formula:

$f1(x)=f(x0)+f(x1)−f(x0)x1−x0(x−x0)$

And, formula for percentage relative error is,

$εt=|True value−Approximate valueTrue value|×100%$

Calculation:

Consider the values, $log8=0.9030900$ and $log12=1.0791812$ .

Here, $x0=8$ and $x1=12$ . Therefore,

$f(x0)=0.9030900f(x1)=1.0791812$

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

$f1(10)=0.9030900+1.0791812−0.903090012−8(10−8)=0.9030900+0.17609124×2=0.9030900+0.0880456=0.9911356$

Now, the true value of $log10=1$ . Therefore, relative percentage error is,

$εt=1−0.99113561×100%=0.0088644×100%=0.88644%$

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

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

To determine

To calculate: The common logarithmic of $10$ by the linear interpolation between $log8=0.9030900$ and $log12=1.0791812$ . Also, find the percentage relative error based on the true value.

Answer 13

Answer to Problem 1P

Solution:

The common logarithmic of $10$ by the linear interpolation between $log8$ and $log12$ is $0.9911356$ with relative percentage error of $0.88644%$ .

Answer 14

Explanation of Solution

Given Information:

The values,

$log8=0.9030900log12=1.0791812$

Formula used:

Linear interpolation formula:

$f1(x)=f(x0)+f(x1)−f(x0)x1−x0(x−x0)$

And, formula for percentage relative error is,

$εt=|True value−Approximate valueTrue value|×100%$

Calculation:

Consider the values, $log8=0.9030900$ and $log12=1.0791812$ .

Here, $x0=8$ and $x1=12$ . Therefore,

$f(x0)=0.9030900f(x1)=1.0791812$

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

$f1(10)=0.9030900+1.0791812−0.903090012−8(10−8)=0.9030900+0.17609124×2=0.9030900+0.0880456=0.9911356$

Now, the true value of $log10=1$ . Therefore, relative percentage error is,

$εt=1−0.99113561×100%=0.0088644×100%=0.88644%$

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

Answer 15

(b)

Expert Solution

To determine

To calculate: The common logarithmic of $10$ by the linear interpolation between $log9=0.9542425$ and $log11=1.0413927$ . 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 $log9$ and $log11$ is $0.9978176$ with relative percentage error of $0.218%$ .

Explanation of Solution

Given Information:

The values,

$log9=0.9542425log11=1.0413927$

Formula used:

Linear interpolation formula:

$f1(x)=f(x0)+f(x1)−f(x0)x1−x0(x−x0)$

And, formula for percentage relative error is,

$εt=|True value−Approximate valueTrue value|×100%$

Calculation:

Consider the values, $log9=0.9542425$ and $log11=1.0413927$ .

Here, $x0=9$ and $x1=11$ . Therefore,

$f(x0)=0.9542425f(x1)=1.0413927$

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

$f1(10)=0.9542425+1.0413927−0.954242511−9(10−9)=0.9542425+0.08715022×1=0.9542425+0.0435751=0.9978176$

Now, the true value of $log10=1$ . Therefore, relative percentage error is,

$εt=|1−0.99781761|×100%=0.0021824×100%=0.218%$

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

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

To determine

To calculate: The common logarithmic of $10$ by the linear interpolation between $log9=0.9542425$ and $log11=1.0413927$ . Also, find the percentage relative error based on the true value.