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Chapter 4: Problem 15

Find a number \(x\) such that \(e^{3 x-1}=2\).

Short Answer

Expert verified
The number \(x\) that satisfies the equation \(e^{3x-1} = 2\) is \(x = \frac{\ln(2) + 1}{3}\).

Step by step solution

01

Take the natural logarithm of both sides

We have the equation \(e^{3x-1} = 2\). To eliminate the exponent, we will take the natural logarithm of both sides. The natural logarithm of a number is the exponent to which e must be raised to produce that number. So, \[\ln(e^{3x-1}) = \ln(2)\]
02

Use the logarithmic property to simplify

Using the logarithmic property \(\ln(a^b) = b \cdot \ln(a)\), we can simplify the left-hand side of the equation: \[(3x-1)\ln(e) = \ln(2)\] The natural logarithm of \(e\) is 1, as \(e^1 = e\). Therefore, the equation becomes: \[3x-1 = \ln(2)\]
03

Solve for \(x\)

Now, we need to isolate \(x\). To do this, we will first add 1 to both sides of the equation: \[3x = \ln(2) + 1\] Next, divide both sides by 3 to find the value of \(x\): \[x = \frac{\ln(2) + 1}{3}\] So, the number \(x\) that satisfies the given equation is \(x = \frac{\ln(2) + 1}{3}\).

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