Problem 123 - College Algebra Essentials

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 is \(x = 82\), whereas for the second equation, it is \(x = 5\). The first equation is solved by converting the logarithm to an exponent, whereas the second equation is solved by utilizing a property of logarithms that allows for the argument of the logarithms to be set equal to each other when the bases and the results of the logarithms are the same.

Step 1: Solve the first equation

The equation is \(\log _{3}(x-1)=4\). Convert this logarithmic equation to an exponential equation. By the definition of logarithms, if \(\log _{b}(a) = c\) then \(b^c = a\). Therefore, by using this property, the equation becomes \(3^4=x-1\), which simplifies to \(81 = x-1\). By adding 1 to both sides, we get \(x = 82\).

Step 2: Solve the second equation

The equation is \(\log _{3}(x-1)=\log _{3} 4\). According to the property of logarithms, if \(\log _{b} M = \log _{b} N\), then \(M = N\). Applying this property, the equation simplifies to \(x - 1 = 4\). By adding 1 to both sides, the equation becomes \(x = 5\).

Step 3: Summary of Differences

For the first equation, the logarithmic equation is converted to an exponential equation to solve for 'x'. In the second equation, an identity (property) of logarithms is used to solve for 'x' directly. Thus, although both equations appear similar, they are solved using different approaches due to the different structures of the equations.