Question

1. the function ( k ) is given by ( k(x)=2 e^{3 x}-e ). solve ( k(x)=7 e ) for values of ( x ) in the domain of ( k ).

Explanation:

Step1: Substitute \(k(x)\) into the equation

Given \(k(x)=2e^{3x}-e\) and \(k(x) = 7e\), we substitute to get \(2e^{3x}-e=7e\).

Step2: Simplify the equation

Add \(e\) to both sides: \(2e^{3x}=7e + e\), so \(2e^{3x}=8e\). Then divide both sides by \(2e\): \(\frac{2e^{3x}}{2e}=\frac{8e}{2e}\), which simplifies to \(e^{3x - 1}=4\).

Step3: Take the natural logarithm of both sides

Using the property \(\ln(e^{a})=a\), if \(e^{3x - 1}=4\), then \(\ln(e^{3x - 1})=\ln(4)\). So \(3x-1=\ln(4)\).

Step4: Solve for \(x\)

Add \(1\) to both sides: \(3x=\ln(4)+1\). Then divide by \(3\): \(x=\frac{\ln(4)+1}{3}\). Since \(\ln(4)=\ln(2^{2}) = 2\ln(2)\), we can also write \(x=\frac{2\ln(2)+1}{3}\).

Answer:

\(x=\frac{\ln(4)+1}{3}\) (or \(x=\frac{2\ln(2)+1}{3}\))