Problem 79 - 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. $$\log (3 x-3)=\log (x+1)+\log 4$$

Step-by-Step Solution

Answer

The equation has no solution.

Step 1: Combine Logs

Start by combining the right hand side logs using the rule \(\log_a(x) + \log_a(y) = \log_a(xy)\). Thus, the equation turns into \(\log(3x - 3) = \log(4(x+1))\).

Step 2: Apply Cancellation Rule

Next, apply the cancellation rule of logarithms since the logarithms have the same base: \(a = b \Rightarrow \log_a = \log_b\). This gives the equation \(3x - 3 = 4x + 4\).

Step 3: Solve for x

Now, isolate the variable \(x\) to find the solution. Subtract \(3x\) from each side to get \(x = -7\).

Step 4: Checking our solution

Verify that the solution doesn't make any expression undefined. However, It is seen that substituting this value of \(x\) into the original equation yields a negative value for the argument of the logarithm, which is not possible since the domain of logarithm function is from 0 to \(\infty\). Hence, the solution \(x = -7\) is rejected.