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Chapter 3: Problem 11

Determine whether each \(x\) -value is a solution (or an approximate solution) of the equation. \(\ln (2 x+3)=5.8\) (a) \(x=\frac{1}{2}(-3+\ln 5.8)\) (b) \(x=\frac{1}{2}\left(-3+e^{5.8}\right)\) (c) \(x \approx 163.650\)

Short Answer

Expert verified
(a) The first value of \(x=\frac{1}{2}(-3+\ln 5.8)\) is not a solution. (b) The second value of \(x=\frac{1}{2}\left(-3+e^{5.8}\right)\) is also not a solution. (c) The third value of \(x \approx 163.650\) is approximately a solution.

Step by step solution

01

Substitute the first \(x\)-value

Substitute \(x=\frac{1}{2}(-3+\ln 5.8)\) into the equation, \(\ln (2 x+3)=5.8\), and see if the equation holds true.
02

Substitute the second \(x\)-value

Substitute \(x=\frac{1}{2}\left(-3+e^{5.8}\right)\) into the equation, \(\ln (2 x+3)=5.8\), and see if the equation holds true.
03

Substitute the third \(x\)-value

Substitute \(x \approx 163.650\) into the equation, \(\ln (2 x+3)=5.8\), and see if the equation holds true. In this case, since we have an approximate value, we are primarily checking if the left side of the equation is approximately equal to 5.8.
04

Verify the results

Verify whether the left side of the equation corresponds to the right side after each substitution. If the equation holds true for a given \(x\)-value, then that \(x\)-value is either a solution or an approximate solution of the equation.

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