Problem 70 - 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. $$\log _{4}(x+2)-\log _{4}(x-1)=1$$

Step-by-Step Solution

Answer

Therefore, the exact solution to the equation \(\log _{4}(x+2)-\log _{4}(x-1)=1) is \(x = 2\).

Step 1: Simplify logarithmic expression

By applying the rule of logarithmic subtraction, we would have: \(\log _{4}(x+2) - \log _{4}(x-1) = \log _{4}\left(\frac{x+2}{x-1}\right)\). Therefore, the given equation simplifies to: \(\log _{4}\left(\frac{x+2}{x-1}\right) = 1\)

Step 2: Solve for x

To get rid of the logarithm and find the value of \(x\), we employ the definition of the logarithm (base \(4\)): \[ 4^1 = \frac{x+2}{x-1} \] This simplifies to: \[ 4x - 4 = x + 2 \] Obtaining: \[ 3x = 6 \] thus: \[ x = 2 \]

Step 3: Verify domain restrictions

We substitute \(x = 2\) back into the original logarithmic expressions to confirm it falls within the domain: \[ \log _{4}(2+2) = \log _{4}(4) = 1/2 \] and \(\log _{4}(2-1) = \log _{4}(1) = 0\). As both results do not lead to a negative log argument, \(x = 2\) is our solution.

Step 4: Decimal approximation

This step is not necessary in our case because the solution is already an exact solution. We do not need to round it up or down.