Problem 78 - 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. $$\log (5 x+1)=\log (2 x+3)+\log 2$$

Step-by-Step Solution

Answer

The solution to the logarithmic equation is \(x = 5\).

Step 1: Combine the logarithms

Because we can add logarithms to multiply their contents, the right side of the equation can be rewritten as \(\log [(2x +3) \cdot 2]\), which simplifies to \(\log [4x +6]\). The equation now becomes: \(\log (5x +1) = \log (4x +6)\).

Step 2: Equalize the arguments

If the logarithm (with the same base) of two expressions are equal, then the expressions themselves must also be equal. Therefore, we can say that \(5x +1 = 4x +6\).

Step 3: Solve for \(x\)

Rearranging the last equation gives \(x = 5\).

Step 4: Confirm the solution is valid

We substitute \(x = 5\) back into the original logarithmic expressions, (both \(5x + 1\) and \(2x + 3)\) to make sure that the results are defined and greater than zero. For \(x = 5\), \(5x + 1 = 26\) and \(2x + 3 = 13\), all of which are positive, indicating \(x = 5\) is indeed a valid solution.