Problem 90 - College Algebra Essentials

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 logarithmic equation is \(x = 3\).

Step 1: Apply log subtraction rule

Apply the log subtraction rule to combine the two logarithms on the left. The difference of two logarithms is equal to the logarithm of the quotient for the same base. Hence, the equation becomes \[\log _{2}\left(\frac{x-1}{x+3}\right)=\log _{2}\left(\frac{1}{x}\right)\].

Step 2: Equating the Arguments

Since the two logarithms with the same base are equal, their corresponding numbers (arguments) must also be equal. So we can set \(\frac{x-1}{x+3}\) equal to \(\frac{1}{x}\). Thus, the equation becomes \[\frac{x-1}{x+3}=\frac{1}{x}\].

Step 3: Cross multiplication.

Cross multiply to clear the fractions. The step results in \[x * (x-1) = (x+3) * 1\]. Simplifying that we get \(x^2-x = x+3.\)

Step 4: Simplify the equation

Simplify the equation to its standard quadratic form. Rearrange the equation such that all terms are on one side of the equation. This yields \(x^2-2x-3=0\). This can be solved by factoring.

Step 5: Solve for x

Factor the quadratic expression and set it equal to zero to solve for \(x\). This yields the factored form \((x-3)(x+1) = 0\). This results in \(x = 3\) and \(x = -1\).

Step 6: Exclude invalid solutions

The solution \(x = -1\) is not a valid solution because if it is plugged back into the original logarithmic expressions it would result in taking the logarithm of a negative number, which is undefined. Therefore, \(x = -1\) is rejected.