\chapter{Exponentials and logarithms}
\section{Exponential equations}
Exponential equations are equations where the unknown is in the exponent.
First we discuss some easy cases.
\subsection{Equating exponents}
If each side of the equation is an
exponential with the same base, then we can just equate the
exponents:
\begin{example}
Suppose we want to solve the equation
\[e^{x-3}=e^{2x+5}.\]
The base is the same, so it's the same exponential function \(y=e^x\) on both sides
of the equation. We know from the graph of \(e^x\) that the function is always increasing
and so it is one-to-one. So there can only be one input for a given output, and to solve
the equation we just equate the two input values:
\[x-3 = 2x + 5.\]
This is easily solved for \(x\), and we find \(x=-8\).
\end{example}
Sometimes we can get the same base by re-writing one or both sides.
\begin{example}
We want to solve
\[3^{x-1}=9.\]
The right side is a power of \(3\), so we can write
\[3^{x-1}=3^2.\]
Now both sides are exponentials with the same base, and we equate the exponents as before:
\[x-1 = 2\]
and we find the solution \(x=3\).
\end{example}
\begin{example}
We now solve \(4^x=8^{x-1}\). We cannot re-write \(4\) as a power of \(8\) or \(8\) as a
power of \(4\). But both \(4\) and \(8\) are powers of \(2\): \(4=2^2\) and \(8=2^3\). So we can re-write the equation as
\[\left(2^2\right)^x = \left(2^3\right)^{x-1}\]
Note the use of parentheses. Now use the property of exponents \(\left(x^n\right)^m = x^{nm}\):
\[2^{2x} = 2^{3(x-1)}\]
Once again we have two exponentials with the same base, so we equate the exponents:
\[2x=3(x-1)\]
Solving this equation we find \(x=3\).
\end{example}
We may need to be creative when thinking about the base. Remember that exponent zero
will give \(1\) no matter what the base is.
\begin{example}
We want to solve \(5^{4x+1}=1\). We re-write \(1\) as \(5^0\):
\[5^{4x+1}=5^0.\]
Now both sides are exponentials with the same base, and we equate the exponents:
\[4x+1=0\]
and solving we find \(x=-1/4\).
\end{example}
\subsection{Using the translation formula}
Sometimes to solve an exponential equation it is enough to use the translation formula from
Unit 3.2:
\[b^x=y \hspace{2ex} \mbox{is equivalent to} \hspace{2ex} \log_b y = x\]
\begin{example}
We want to solve the equation
\[e^{2x+3}= 5.\]
Using the translation formula, this becomes
\[\ln 5 = 2x+3\]
and when we solve it for \(x\) we find
\[x=\frac{\ln 5 -3}{2}.\]
\end{example}
\subsection{Taking \(\ln\) of both sides}
There are cases when the bases are different and cannot be re-written as a common base, and
the translation formula is not enough. In this situation there is a method that will always work,
even though it may require more steps: take \(\ln \) of both sides.
%This is similar to when we want to solve a quadratic
%equation: if it can be factored, then that is the quickest way to do it, but if it cannot, we can always
%solve it with the quadratic formula, that takes more work.
\begin{example}
Suppose we want to solve \[5^x = 3^{2x+1}.\]
Clearly we cannot re-write \(3\) and \(5\) using a common base, and the translation formula
can't be used because there are two different bases involved. So we take \(\ln \) of both sides:
\[\begin{array}{rcll}
5^x & = & 3^{2x+1} & \mbox{The given equation}\\[1.5ex]
\ln 5^x & = & \ln 3^{2x+1} & \mbox{Take \(\ln\) of both sides}\\[1.5ex]
x \ln 5 & = & (2x+1) \ln 3 & \mbox{Use property (C) to bring down the exponents}\\
&&& \mbox{Don't forget to include the parentheses}\\[2ex]
x\ln 5 & = & 2x\ln 3 +\ln 3 & \mbox{Expand the right side\\
Treat \(\ln 3\) like any other number}\\[2ex]
x \ln 5 \mathbf{-2x\ln 3} & = & 2x \ln 3 + \ln 3 \mathbf{-2x\ln 3} &
\mbox{Remember that we are looking for \(x\)}\\
&&& \mbox{So subtract \(2x \ln 3\) from both sides}\\[2ex]
x\ln 5 -2x \ln 3& = & \cancel{2x\ln 3} + \ln 3 \cancel{-2x\ln 3}& \mbox{Simplify}\\[1.5ex]
x\ln 5 -2x \ln 3& = & \ln 3 & \\[1.5ex]
x(\ln 5 - 2 \ln 3) & = & \ln 3 & \mbox{Factor \(x\)}\\[1.5ex]
\displaystyle \frac{x(\ln 5 -2\ln 3)}{\mathbf{\ln 5 -2 \ln 3}} & = &
\displaystyle \frac{\ln 3}{\mathbf{\ln 5 -2 \ln 3}} & \mbox{\(\ln 5 -2\ln 3\) is just a number. So divide both sides by it}\\[1.5ex]
\displaystyle \frac{x\cancel{(\ln 5 -2\ln 3)}}{\cancel{\ln 5 -2 \ln 3}} & = &
\displaystyle \frac{\ln 3}{\ln 5 -2 \ln 3} & \mbox{Simplify}\\[1.5ex]
x & = & \displaystyle \frac{\ln 3}{\ln 5 -2 \ln 3} & \mbox{Find the answer}
\end{array}
\]
\end{example}
If a translation formula is enough to solve an equation, but the base of the logarithm
is not \(10\) or \(e\), we prefer to solve the equation by taking \(\ln\) of both sides. This is
to avoid having the answer with a logarithm in a base that is not \(10\) or \(e\) (and that
could not be found with a calculator).
\begin{example}
We want to solve the equation
\[5^{3x}-2=8.\]
First we add \(2\) to both sides, to get the exponential on the left by itself:
\[5^{3x} =10\]
Now we could solve the equation by using the translation formula: \(5^y=x\) is equivalent
to \(\log_5 x =y\). But then we would find an answer that uses \(\log_5\). Instead, we take \(\ln\)
of both sides:
\[\ln 5^{3x}=\ln 10.\]
Using property (C), we find
\[3x \ln 5 = \ln 10,\]
and dividing both sides by \(3 \ln 5\) we find the answer:
\[x=\frac{\ln 10 }{3\ln 5}\]
\end{example}
Taking the natural logarithm \(\ln\) of both sides will always work. But make sure you apply
the properties of the logarithms correctly after that step.
\begin{example}
We want to solve
\[3 e^{1-3x}= 2^{x+1}.\]
On the left side of the equation, the exponent \(1-3x\) belongs only to \(e\). The number \(3\)
is just multiplied by the exponential \(e^{1-3x}\).
\[\begin{array}{rcll}
3 e^{1-3x} & = & 2^{x+1} & \mbox{The given equation}\\[1.5ex]
\ln \left(3 e^{1-3x}\right) & = & \ln2^{x+1} & \mbox{Take \(\ln\) of both sides}\\[1.5ex]
\ln 3+ \ln e^{1-3x} & = & \ln 2^{x+1} & \mbox{We cannot use property (C) yet.}\\
&&& \mbox{Instead use property (A) first}\\[2ex]
\ln 3 + 1-3x & = & (x+1) \ln 2 & \mbox{Use the simplification formula \(\ln e^y=y\)}\\
&&& \mbox{on the left and property (C) on the right}\\[2ex]
\ln 3+1 -3x & = & x \ln 2 + \ln 2 &
\mbox{Expand the right side}\\[1.5ex]
\ln 3+1 -3x \mathbf{-x \ln 2}&= & x \ln 2 +\ln 2 \mathbf{ -x \ln 2} & \mbox{Subtract \(x \ln 2\) from both sides}\\[1.5ex]
\ln 3 + 1 -3x -x \ln 2& = & \ln 2 & \mbox{Simplify}\\[1.5ex]
\ln 3 +1 -3x -x \ln 2 \mathbf{-\ln 3 -1} & = & \ln 2 \mathbf{-\ln 3 -1} &\mbox{Subtract \(\ln 3 +1\) from both sides} \\[1.5ex]
-3x - x \ln 2 & = & \ln 2-\ln 3 -1 & \mbox{Simplify}\\[1.5ex]
3x + x \ln 2 & = & -\ln 2+\ln 3 +1 & \mbox{multiply both sides by \(-1\)}\\[1.5ex]
x(3+\ln 2) & = & \ln 3 +1-\ln 2 & \mbox{Factor \(x\) and write the positive terms first}\\[1.5ex]
\displaystyle \frac{x(3+\ln 2)}{\mathbf{(3+\ln 2)}} & = &
\displaystyle \frac{\ln 3 +1-\ln 2}{\mathbf{(3+\ln 2)}} & \mbox{Divide both sides by \((3+\ln 2)\)}\\[2ex]
\displaystyle \frac{\cancel{(3+\ln 2)}x}{\cancel{(3+\ln 2)}} & = &
\displaystyle \frac{\ln 3 +1-\ln 2}{(3+\ln 2)} & \mbox{Simplify}\\[2ex]
x & = & \displaystyle \frac{ \ln 3 +1-\ln 2}{3+\ln 2} &\mbox{Find the answer}
\end{array}
\]
\end{example}
Problems
\problem
Solve the exponential equations.
\[
\begin{array}{lll}
\mbox{a. } 3^{3x-4}=3^{1-3x}& \mbox{b. } 2^{5x-2}=8& \mbox{c. } 27^{2x+1}=9^{1-x} \\[1ex]
\mbox{d. } 4^{3-2x}=1& \mbox{e. } 7=10^{2x-5}&
\end{array}
\]
\begin{sol}
\begin{enumerate}
\item
\(3^{3x-4}=3^{1-3x}\)
Both sides are exponentials with base 3. So we equate the exponents and solve:
\[\begin{array}{rcll}
3x-4 & = & 1-3x & \mbox{equate exponents}\\[1.5ex]
3x - 4 \mathbf{ +3x} & = & 1-3x \mathbf{+3x} & \mbox{add \(3x\) to both sides}\\[1.5ex]
6x-4 & = & 1 & \mbox{simplify}\\[1.5ex]
6x -4 \mathbf{+4} & = & 1 \mathbf{+4} & \mbox{add \(4\) to both sides}\\[1.5ex]
6x & = & 5 & \mbox{simplify}\\[1.5ex]
\displaystyle \frac{6x}{\mathbf{6}} & = & \displaystyle \frac{5}{\mathbf{6}} &
\mbox{divide by \(6\)}\\[1.5ex]
x & = & \displaystyle \frac{5}{6} & \mbox{simplify}
\end{array}
\]
\item \
\(2^{5x-2}=8\)
\[\begin{array}{rcll}
2^{5x-2} & = & 2^3 & \mbox{re-write \(8\) as \(2^3\)}\\[1.5ex]
5x-2 & = & 3 & \mbox{equate exponents}\\[1.5ex]
5x -2 \mathbf{+2} & = & 3\mathbf{+2} & \mbox{add \(2\) to both sides}\\[1.5ex]
5x & = & 5 & \mbox{simplify}\\[1.5ex]
\displaystyle \frac{5x}{\mathbf{5}} & = & \displaystyle \frac{5}{\mathbf{5}} &
\mbox{divide both sides by \(5\)}\\[1.5ex]
x & = & 1 & \mbox{simplify}\\[1.5ex]
\end{array}
\]
\item \
\(27^{2x+1}=9^{1-x}\)
\[\begin{array}{rcll}
\left(3^3\right)^{2x+1} & = & \left(3^2\right)^{1-x} & \mbox{re-write both sides
as powers of \(3\)}\\[1.5ex]
3^{3(2x+1)} & = & 3^{2(1-x)} & \mbox{use properties of the exponents}\\[1.5ex]
3(2x+1) & = & 2(1-x) & \mbox{equate exponents}\\[1.5ex]
6x+3 & = & 2-2x & \mbox{expand}\\[1.5ex]
6x+3 \mathbf{+2x} & = & 2 -2x \mathbf{+2x} &
\mbox{add \(2x\) to both sides}\\[1.5ex]
8x +3& = & 2 & \mbox{simplify}\\[1.5ex]
8x + 3 \mathbf{-3} & = & 2 \mathbf{-3} & \mbox{subtract \(3\) from both sides}\\[1.5ex]
8x & = & -1 & \mbox{simplify} \\[1.5ex]
\displaystyle \frac{8x}{\mathbf{8}} & = & \displaystyle \frac{-1}{\mathbf{8}}
& \mbox{divide both sides by \(8\)}\\[1.5ex]
x & = &\displaystyle -\frac{1}{8} & \mbox{simplify}
\end{array}
\]
\item
\(4^{3-2x}=1\)
\[\begin{array}{rcll}
4^{3-2x} & = & 4^0 & \mbox{re-write \(1\) as \(4^0\)}\\[1.5ex]
3-2x & = & 0 & \mbox{equate exponents}\\[1.5ex]
3-2x \mathbf{-3} & = & 0\mathbf{-3} & \mbox{subtract \(3\) from both sides}\\[1.5ex]
-2x & = & -3 & \mbox{simplify}\\[1.5ex]
\displaystyle \frac{-2x}{\mathbf{-2}} & = & \displaystyle \frac{-3}{\mathbf{-2}} &
\mbox{divide both sides by \(-2\)}\\[1.5ex]
x& = & \displaystyle \frac{3}{2} & \mbox{simplify}
\end{array}
\]
\item \ \(7=10^{2x-5}\)
\[\begin{array}{rcll}
7 & = & 10^{2x-5} & \mbox{The given equation}\\[1.5ex]
2x-5 & = & \log 7 & \mbox{Use the translation formula:}\\
& & & \mbox{\(10^x=y\) is equivalent to \(x=\log y\)}\\[1.5ex]
2x -5 \mathbf{+5} & = &\log 7 \mathbf{+5} & \mbox{Add \(5\) to both sides}\\[1.5ex]
2x & = & 5+\log 7 & \mbox{Simplify}\\[1.5ex]
\displaystyle \frac{2x}{\mathbf{2}} & = &\displaystyle \frac{5+\log 7}{\mathbf{2}} &
\mbox{Divide both sides by \(2\)}\\[1.5ex]
x& = & \displaystyle \frac{5+\log 7}{2} & \mbox{Simplify}\\[1.5ex]
\end{array}
\]
\end{enumerate}
\end{sol} \mproblem
Solve the exponential equations.
\[
\begin{array}{ccccc}
\mbox{a. } 2^{3x+1}=16& \mbox{b. } e^{2x-3}=e^{1-x}& \mbox{c. } 27^{x+1}=9^{2x-3}& \mbox{d. } 3^{5x+3}=1
& \mbox{e. } e^{5x+3}=4
\end{array}
\]
\problem
Solve the equations:
\[
\begin{array}{ccc}
\mbox{a. } 3^{x+4}=6& \mbox{b. } 5^{2x-3}=2^{1-x}& \mbox{c. } 4^x =2 e^{2x+1}
\end{array}
\]
\begin{sol}
\begin{enumerate}
\item \
\[\begin{array}{rcll}
3^{x+4}& = & 6 & \mbox{The given equation}\\[1ex]
\ln 3^{x+4} & =\ln 6 & & \mbox{The translation formula would use \(\log_3\).}\\
&&& \mbox{Instead, we take \(\ln\) of both sides}\\[2ex]
(x+4) \ln 3 & = \ln 6 & & \mbox{Use property (C)}\\[1.5ex]
x \ln 3 + 4\ln 3 & = & \ln 6 & \mbox{Expand the left side}\\[1.5ex]
x \ln 3 + 4\ln 3 \mathbf{-4\ln 3} & = & \ln 6 \mathbf{-4 \ln 3} &
\mbox{Subtract \(4 \ln 3\) from both sides}\\[1.5ex]
x \ln 3 & = & \ln 6 - 4 \ln 3 & \mbox{Simplify}\\[1.5ex]
\displaystyle \frac{x \ln 3}{\mathbf{\ln 3}}&= & \displaystyle \frac{\ln 6 -4 \ln 3}{\mathbf{\ln 3}} & \mbox{Divide both sides by \(\ln 3\)}\\[1.5ex]
x & = & \displaystyle \frac{\ln 6 -4 \ln 3}{\ln 3} & \mbox{Simplify}
\end{array}
\]
\item \
\[\begin{array}{rcll}
5^{2x-3} & = 2^{1-x} & & \mbox{The given equation}\\[1.5ex]
\ln 5^{2x-3} & = & \ln 2^{1-x} & \mbox{Take \(\ln\) of both sides}\\[1.5ex]
(2x-3) \ln 5 & = & (1-x) \ln 2 & \mbox{Use property (C)}\\[1.5ex]
2x \ln 5 -3 \ln 5& = & \ln 2 - x\ln 2 & \mbox{Expand}\\[1.5ex]
2x \ln 5 - 3\ln 5 \mathbf{+x\ln 2}& = &\ln 2 -x \ln 2 \mathbf{+x \ln 2} &
\mbox{Add \(x \ln 2\) to both sides}\\[1.5ex]
2x\ln 5 -3 \ln 5 +x\ln 2&= & \ln 2 & \mbox{Simplify}\\[1.5ex]
2x\ln 5 -3 \ln 5 +x\ln 2 \mathbf{+3 \ln 5}& = & \ln 2 \mathbf{+3 \ln 5} &
\mbox{Add \(3\ln 5\) to both sides}\\[1.5ex]
2x\ln 5 +x\ln 2& = & \ln 2 +3 \ln 5 &\mbox{Simplify} \\[1.5ex]
x(2\ln 5+\ln 2)& = & \ln 2 +3 \ln 5 & \mbox{Factor \(x\)}\\[1.5ex]
\displaystyle \frac{x(2\ln 5+\ln 2)}{2\ln 5+\ln 2} & = & \displaystyle \frac{\ln 2 +3 \ln 5}{2\ln 5+\ln 2} & \mbox{Divide both sides by \(2\ln 5+\ln 2\) }\\[2ex]
x & = & \displaystyle \frac{\ln 2 +3 \ln 5}{2\ln 5+\ln 2} & \mbox{Simplify}
\end{array}
\]
\item \
\[\begin{array}{rcll}
4^x & = & 2 e^{2x+1} & \mbox{The given equation}\\[1.5ex]
\ln 4^x & = & \ln \left(2e^{2x+1}\right) & \mbox{Take \(\ln\) of both sides}\\[1.5ex]
x \ln 4& = & \ln 2 + \ln e^{2x+1} & \mbox{Use property (C) on the left}\\
&&& \mbox{and property (A) on the right}\\[2ex]
x \ln 4 & = & \ln 2 + 2x+1 & \mbox{Use the simplification formula \(\ln e^x = x\) }\\[1.5ex]
x \ln 4 \mathbf{-2x}& = & \ln 2 +2x +1 \mathbf{-2x} &
\mbox{Subtract \(2x\) from both sides}\\[1.5ex]
x(\ln 4-2)&= &\ln 2+1 & \mbox{Simplify, and factor \(x\)}\\[1.5ex]
\displaystyle \frac{x(\ln 4-2)}{\mathbf{\ln 4-2}}& = & \displaystyle \frac{\ln 2 +1}{\mathbf{\ln 4-2}} & \mbox{Divide both sides by \(\ln 4 -2\)}\\[2ex]
x& = & \displaystyle \frac{\ln 2 +1}{\ln 4-2} &\mbox{Simplify} \\[1.5ex]
\end{array}
\]
\end{enumerate}
\end{sol} \mproblem
Solve the equations:
\[
\begin{array}{ccc}
\mbox{a. } 6^{2x}-5=7& \mbox{b. } 3^{4x-1}=7^{2-x}& \mbox{c. } 4e^{3x-5} = 2^x
\end{array}
\]