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

Step-by-Step Solution

Answer

No feasible solution, because the obtained solution is out of the domain of original equation's logarithmic expressions.

Step 1: Combine Logs

You have two logs on the right side of equation which you can combine into one log via the rule of logarithms \( \log(a)+\log(b)= \log(a*b)\). This will give: \(\log (2x-1) = \log(3*(x+3))\).

Step 2: Remove the Logarithm

Since the base of the logs is same, you can remove the logs via the rule of equality \( \log(a)=\log(b)\) implies \( a=b\). Hence, the equation now looks like this: \(2x -1 = 3(x + 3)\)

Step 3: Solve for x

Simplify the equation to find the value of x: First distribute 3 over \(x + 3\), yields \(2x - 1 = 3x + 9\). Then, solving this equation for x yields \(x = -10\)

Step 4: Verify the Solution

You have to make sure the solution, \(x = -10\), is valid in the original equation. Since the domain of logarithmic function, \(log(x)\), is \( x>0\), with \(x = -10\), both \(2*(-10) - 1 < 0\) and \(-10 + 3 < 0\). Both these values are out of domain of the logarithm function. Hence, -10 must be rejected.