Problem 46 - A Graphical Approach to Precalculus with Limits

Question

Solve each logarithmic equation. Express all solutions in exact form. Support your solutions by using a calculator. $$1-4 \ln (2 x-1)=-5$$

Step-by-Step Solution

Answer

The solution is \(x = \frac{e^{\frac{3}{2}} + 1}{2}\).

Step 1: Isolate the logarithmic term

Start by isolating the logarithmic term on one side of the equation. Begin with the equation given as \(1 - 4 \ln(2x - 1) = -5\). Add 4 to both sides of the equation to get \(-4 \ln(2x - 1) = -6\). Next, divide every term by -4 to isolate the logarithm: \(\ln(2x - 1) = \frac{6}{4}\). Simplify \(\frac{6}{4}\) to \(\frac{3}{2}\).

Step 2: Convert the logarithmic equation to exponential form

Since \(\ln(2x - 1) = \frac{3}{2}\), you can write this in exponential form. Recall that if \(\ln(a) = b\), then \(e^b = a\). Therefore, \(2x - 1 = e^{\frac{3}{2}}\).

Step 3: Solve for the variable

Now, solve for \(x\) by isolating it. Start with \(2x - 1 = e^{\frac{3}{2}}\). Add 1 to both sides to get \(2x = e^{\frac{3}{2}} + 1\). Finally, divide each term by 2 to solve for \(x\): \(x = \frac{e^{\frac{3}{2}} + 1}{2}\).

Step 4: Verify the solution using a calculator

Use a calculator to confirm the value of \(x\). Calculate \(e^{\frac{3}{2}}\) to ensure that you understand the expression. Then, solve \(e^{\frac{3}{2}} + 1\), divide by 2 and compare the result to log checks in the original equation to ensure correctness.

Key Concepts

Isolating Logarithmic Terms

When solving a logarithmic equation, one of the initial steps is to ensure that the logarithmic term stands alone on one side of the equation. This is imperative for simplifying complex equations. Consider the equation given: \(1 - 4 \ln(2x - 1) = -5\). The goal here is to isolate \(\ln(2x - 1)\). Start this process by eliminating constants and coefficients surrounding the logarithmic expression. Here’s how you can achieve this:

• First, add 4 to both sides: \(-4 \ln(2x - 1) = -6\).
• Next, divide each component by -4 to release the \(\ln(2x - 1)\) term: \(\ln(2x - 1) = \frac{6}{4}\).
• Simplify \(\frac{6}{4}\) to \(\frac{3}{2}\).

Now, the equation is nicely unraveled to \(\ln(2x - 1) = \frac{3}{2}\), allowing smooth progress to an exponential form conversion.

Exponential Form

Transitioning a logarithmic equation to exponential form can be seen as a bridge to uncovering the variable. This ties back to the fundamental understanding of what a logarithm represents in relation to exponents. Given \(\ln(a) = b\), e.g., \(\ln(2x - 1) = \frac{3}{2}\), the conversion implies \(e^b = a\).So for our equation, this translates to:\[2x - 1 = e^{\frac{3}{2}}\]Through this conversion to exponential form, the expression is transformed into a format where the variable can be approached directly through algebraic manipulation.

Solving for Variables

Once you’ve interpreted the logarithmic equation into its exponential form, the next logical step is to solve for the variable, in this case, \(x\). The equation now reads \(2x - 1 = e^{\frac{3}{2}}\).- Begin by addressing any constants with simple arithmetic. Adjust each side to isolate terms involving the variable.- Add 1 to both sides of the equation to eliminate the negative constant: \(2x = e^{\frac{3}{2}} + 1\).- Finally, divide everything by 2 to solve for \(x\): \[x = \frac{e^{\frac{3}{2}} + 1}{2}\]This provides a clear expression for \(x\) in terms of exact mathematical components. Solving this ensures the variable is appropriately isolated.

Verification with Calculators

Verifying your solution using a calculator complements your analytical efforts, confirming that the steps taken are accurate. It involves calculating the exact numeric value of the expression you derived:- Calculate \(e^{\frac{3}{2}}\) using a calculator to get the precise growth multiple.- Add the resultant number 1 to this value to complete the expression \(e^{\frac{3}{2}} + 1\).- Finally, divide by 2 to verify \(x = \frac{e^{\frac{3}{2}} + 1}{2}\).Check once more by substituting back into the original equation to ensure both sides balance. This approach ensures reliability in your solution and exemplifies the thoroughness necessary in mathematical problem-solving.