Problem 90 - 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. $$\log _{2}(x-1)-\log _{2}(x+3)=\log _{2}\left(\frac{1}{x}\right)$$

Step-by-Step Solution

Answer

The solution to the given logarithmic equation is \(x = 3\).

Step 1: Combine the logarithms

The first step is to combine the logs on the left side of the equation using the properties of logs: \[\log_{2}\left(\frac{(x-1)}{(x+3)}\right) = \log_{2}\left(\frac{1}{x}\right)\]

Step 2: Remove the logs

Since the bases of the logs are the same, we can equate the arguments of the logs (i.e., the expressions inside the logs): \[\frac{(x-1)}{(x+3)} = \frac{1}{x}\]

Step 3: Cross multiply

Cross multiply to get rid of the fractions. That yields: \[x^2 - x = x + 3\]

Step 4: Rearrange and find the roots

Rearrange the terms, form a quadratic equation and solve for \(x\):\[x^2 - 2x - 3 = 0\]This equation can be factored further to:\[(x - 3)(x + 1) = 0\]So, the possible roots are \(x = 3\) and \(x = -1\).

Step 5: Check the domain

Check the obtained values by plugging them back into the original equation. The original expression is defined only if \(x-1\) and \(x+3\) are both positive. Hence the value \(x = -1\) should be rejected because it is not in the domain. Therefore, only \(x = 3\) is the solution.

Step 6: Decimal approximation

The exact solution does not require approximations here, since the solution \(x = 3\) is an integer. However, if requisite, the decimal approximation is also \(3.00\) to two decimal places.