Problem 121 - Introductory and Intermediate Algebra for College Students 4th

Question

Explain the differences between solving \(\log _{3}(x-1)=4\) and \(\log _{3}(x-1)=\log _{3} 4\).

Step-by-Step Solution

Answer

The solution for the first equation \(\log _{3}(x-1)=4\) is \(x=3^4 + 1 = 82\). The solution for the second equation \(\log _{3}(x-1)=\log _{3} 4\) is \(x = 4 + 1 = 5\). The first equation required conversion to an exponential equation for solving, while the second equation could be solved by directly equating the quantities within the logarithms, due to the property of logarithms.

Step 1: Solving the First Equation

At the core, the equation \(\log _{3}(x-1)=4\) tells us that the base 3 raised to the power 4 gives us the quantity \((x-1)\). Hence, our first step is to convert this logarithmic expression into an exponential one, which would be \(3^4=x-1\). Solving this gives us \(x=3^4 + 1\).

Step 2: Solving the Second Equation

On the other hand, the equation \(\log _{3}(x-1)=\log _{3} 4\) implies that the quantity \((x-1)\) and the number 4 are equivalent. This is due to the property of logarithms that if \(\log _{b} a = \log _{b} c\), then \(a = c\). Hence, here we can immediately equate \(x-1\) to 4, and solve to find that \(x = 4 + 1\).

Step 3: Contrasting the Two Methods

As can be observed, the primary difference between solving these two equations lies in the direct application of the property of logarithms in the second equation, as opposed to the conversion from logarithmic to exponential form in the first. Understanding the appropriate situations to apply these different approaches is fundamental in solving logarithm-based problems effectively.