Problem 61 - Introductory and Intermediate Algebra for College Students 4th

Question

Solve each logarithmic equation. Be sure to reject any value of \(x\) that is not in the domain of the original logarithmic expressions. Give the exact answer. Then, where necessary, use a calculator to obtain a decimal approximation, correct to two decimal places, for the solution. $$6+2 \ln x=5$$

Step-by-Step Solution

Answer

The exact answer is \(x = e^{-0.5}\) approximately equal to 0.61 when rounded to the nearest hundredth.

Step 1: Isolate the Logarithmic Term

To isolate the logarithmic term, subtract 6 from both sides of the equation. This gives us \n\n \[2 \ln x = 5 - 6\]. \n \n Simplifying the right side gives us \(2 \ln x = -1\).

Step 2: Divide By Coefficient Of Logarithmic Term

To get rid of the constant coefficient of the logarithmic term, we divide both sides of the equation by 2. This gives us \n \n \[\ln x = -1 / 2 = -0.5\].

Step 3: Solve for x

To express \(x\), we must convert from logarithmic form to exponential form. In this case, the base is \(e\). Using the formula \(a = b^c\) where \(a\) is the argument of the log, \(b\) is the base, and \(c\) is the value of the log, we have \n \n \[x = e^{-0.5}\]. \n \n To get the value of \(x\), we calculate \(e^{-0.5}\). However, before that, we should check if the value is within the domain of the original expression.

Step 4: Check Domain

The base of the natural logarithm function is \(e\), so the domain of any natural logarithm function is \(x > 0\). In this case, \(e^{-0.5}\) is certainly greater than 0, so it's in the domain of the original function.

Step 5: Get decimal value

Now, using a calculator, we find that to two decimal places, \[ e^{-0.5} = 0.61\]