Problem 80 - 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. $$\log (2 x-1)=\log (x+3)+\log 3$$

Step-by-Step Solution

Answer

The equation has no solution as the obtained value of \(x=-10\) does not satisfy the domain constraints of the logarithmic expressions.

Step 1: Combine Logarithms on Right Hand Side

Combine \(\log(x+3)+\log3\) into \(\log(3(x+3))\). This is as per the logarithmic identity \(\log a + \log b = \log(ab)\). The equation then becomes \(\log(2x-1)=\log(3x+9)\).

Step 2: Solve for \(x\)

Since we have a log on each side, we can set the arguments equal to each other \(2x-1=3x+9\). Solving this equation gives \(x=-10\)

Step 3: Check for Domain Constraints

Substituting \(x=-10\) back into original logarithmic expressions \(2x-1\) and \(x+3\) should result in values greater than zero, because the domain of log functions is greater than zero. However substituting \(x=-10\) into both these expressions results in negative values. \(2(-10)-1=-21\) and \(-10+3=-7\) which are both less than zero. Therefore, \(x=-10\) can not be the solution for the original problem, and the equation has no solution.