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Chapter 3: Problem 4

Determine whether each \(x\) -value is a solution (or an approximate solution) of the equation. \(4 e^{x-1}=60\) (a) \(x=1+\ln 15\) (b) \(x=\ln 16\)

Short Answer

Expert verified
Substituting \(x = 1 + \ln 15\) into the equation gives us \(4e^{(1+\ln 15)-1} = 60\), which simplifies to \(4(15) = 60\). Thus, \(x = 1 + \ln 15\) is indeed a solution to the equation. However, substituting \(x = \ln 16\), we get \(4e^{(\ln 16)-1} = 60\), which simplifies to \(4e^{\ln (16/e)} = 60\). Simplifying further to \(4(16/e) = 60\) shows that \(x = \ln 16\) is not a solution to the equation.

Step by step solution

01

Substitute \(x = 1 + \ln 15\) in original equation

Replace \(x\) with \(1+\ln 15\) in the equation, which gives \(4e^{(1+\ln 15)-1} = 60\). Simplify this expression to check if the left side equals 60.
02

Substitute \(x = \ln 16\) in original equation

Replace \(x\) with \(\ln 16\) in the equation, which gives \(4e^{(\ln 16)-1} = 60\). Simplify this equation to check if the two sides are equal.

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