Problem 38 - College Algebra

Question

Solve each logarithmic equation in Exercises \(27-44 .\) Be sure to reject any value of \(x\) that produces the logarithm of a negative number or the logarithm of \(0 .\) $$\log _{2}(x-1)+\log _{2}(x+1)=3$$

Step-by-Step Solution

Answer

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

Step 1: Combine the Logs

Use the property of logarithms that states \(\log_a(x) + \log_a(y) = \log_a(xy)\) to combine the two logarithm expressions on the left side of the equation. This yields \(\log_2((x-1)(x+1))=3\)

Step 2: Convert to exponential form

The logarithmic equation can be expressed in exponential form as \(2^3 = (x-1)(x+1)\)

Step 3: Solve for x

Now we can simply solve for x. First, \(2^3\) equals \(8.\) So, the equation simplifies to \(8=(x-1)(x+1).\) This expands to \(8 = x^2 - 1\). Adding \(1\) to both sides gives \(x^2 = 9\), so \(x = ±3.\)

Step 4: Check solutions

We need to make sure the solutions don't violate the logarithmic expressions' domain. Inserting \(x = 3\) into the original equation, both logarithms are defined, so \(3\) is a solution. Inserting \(x = -3\) into the original equation, the term \(\log_2 (x-1)\) isn't defined, so \(-3\) must be rejected.