\chapter{Exponentials and logarithms}
\section{Logarithmic equations}
Answer the questions.
\begin{enumerate}
\item
What is an important difference between solving an exponential equation
and solving a logarithmic equation?
\item
A student solved the equation \(\log(x+3) +\log (x-5) = \log 2\)
with the following steps:
\[\begin{array}{rcl}
\log(x+3) +\log (x-5) &=& \log 2\\[1.5ex]
(x+3) + (x-5) &=& 2\\[1.5ex]
x+3 + x-5 & = & 2\\[1.5ex]
2x -2 & = & 2\\[1.5ex]
2x & = & 4\\[1.5ex]
x & = & 2
\end{array}
\]
but when she checked the original equation this answer did not work. So she concluded that the
equation has no solution. But that is not the right answer. What went wrong, and what is the right
way to solve the equation?
\item
In Unit 3.5, we discussed the following methods to solve logarithmic equations:
- Equate inputs
- Isolate the logarithm, then use the translation formula
- Use the property of logarithms, then equate inputs
- Use the property of logarithms, then use the translation formula.
For each of the following equations, decide which of the methods i.-iv. is the right one, without
actually solving the equations.
\begin{enumerate}
\item \(\log(2x+7) +\log (5x-1)= \log (3x-1)\)
\item \(3\ln(2x+1)=-1\)
\item \(\log(4x+7) = \log(x-1)\)
\item \(\log_3(2x+3) + \log_3 (x+1) = 1\)
\end{enumerate}
\end{enumerate}