Section 4.5

Remember

• \(a^x=a^y\) if and only if \(x=y\),

• \(\log_a(x)=\log_a(y)\) if and only if \(x=y\),

• \(f(f^{-1}(x))=x\) and \(f^{-1}(f(x))=x\), and

• \(\log_a(a^x)=x\) and \(a^{\log_a(x)}=x\).

Solve Exponential Equations

Example 81

Solve \(8^x=24\)

Solution:

First, we will compose both sides by the natural log function:

\[\ln(8^x)=\ln(24)\to x\ln(8)=\ln(24)\]

Then we will divide both sides by \(\ln(8)\).

\[\frac{x\cancel{\ln(8)}}{\cancel{\ln(8)}}=\frac{\ln(24)}{\ln(8)}\]

Therefore, the solution set is \(\{\frac{\ln(24)}{\ln(8)\}\).

The previous answer can be represented in many different ways.

\[\frac{\ln(24)}{\ln(8)}=\log_8(24)\]

which could lead to:

\[\log_8(24)=\log_8(8\cdot 3)=\log_8(8)+\log_8(3)=1+\log_8(3)\]

or

\[1+\log_8(3) = 1+\frac{\ln(8)}{\ln(3)}\]

It is also important to notice that \(\frac{\ln(24)}{\ln(8)}\ne \ln(3)\).

Example 82

Solve \(e^{4x}e^{x-1}=5e\).

Solution:

First, we will apply the law \(a^m a^n = a^{m+n}\)

\[e^{4x}e^{x-1}=e^{4x+(x-1)}=e^{5x-1}\]

Then we have:

\[e^{5x-1}=5e\]

Next, we will compose both sides by the natural log

\[\begin{align*} \ln(e^{5x-1}) & = \ln(5e)\\ (5x-1)\ln(e) & = \ln(5e)\\ 5x-1 & = \ln(5e) \end{align*}\]

Next, add one to both sides then divide by 5 to solve for \(x\).

\[\begin{align*} 5x-1 & = \ln(5e)\\ 5x & = \ln(5e)+1\\ x & = \frac{\ln(5e)+1}{5} \end{align*}\]

Which can written as

\[\begin{align*} \frac{\ln(5e)+1}{5} & = \frac{\ln(5)+\ln(e)+1}{5}\\ & = \frac{\ln(5)+1+1}{5}\\ & = \frac{\ln(5)+2}{5} \end{align*}\]

The solution set is \(\{\frac{\ln(5)+2}{5}\}\)

Example 83

Solve \(e^{2x}-4e^x+3=0\)

Solution:

First, we will rewrite the equation in the following way

\[\left(e^x\right)^2-4\left(e^x\right)+3=0\]

Then see this is a “quadratic-like” equation. Let \(u=e^x\), then \(u^2=e^{2x}\) and

\[\begin{align*} u^2 - 4u + 3 & = 0\\ (u-3)(u-1) & = 0 \end{align*}\]

With \(u=3\) and \(u=1\) as solutions, we substitute \(u=e^x\).

In the case \(u=3\), we have, \(e^x=3\). After composing both sides of the equation by the natural log we have \(x=\ln(3)\).

In the case \(u=1\), we have \(e^x=1\). After composing both sides of the equation by the natural log we have \(x=\ln(1)=0\).

Therefore, the solution set is \(\{0,\ln(3)\}\).

Example 84

Solve \(e^{2x}+e^x-6=0\).

Solution:

First, rewrite the equation as follows

\[\left(e^x)\right)^2+\left(e^x\right)-6=0\]

Let \(u=e^x\), then \(u^2=e^{2x}\) and

\[\begin{align*} u^2 + u - 6 & = 0\\ (u+3)(u-2) & = 0 \end{align*}\]

For the case \(u=-3\), we have, \(e^x=-3\) which has no solution. Remember \(\ln(-3)\) is undefined.

For the case \(u=2\), we have, \(e^x=2\) which has a solution of \(x=\ln(2)\).

Therefore, the solution set is \(\{\ln(2)\}\).

Solve Logarithmic Equations

When solving log equations it is important to check solutions against the original equation.

Example 85

Solve \(4\ln(x)=36\).

Solution:

First, we will isolate \(\ln(x)\)

\[4\ln(x)=36 \to \ln(x)=\frac{36}{4}\]

Next, we will use the fact that \(e^{\ln(x)}=x\). After composing both sides by the exponential function \(e\) we have

\[\begin{align*} e^{\ln(x)} & = e^{\frac{36}{4}}\\ x & = e^9 \end{align*}\]

Next, we will check the solution.

\[\begin{align*} 4\ln(e^9) & = 36\\ 4\cdot 9 \ln(e) & = 36\\ 36 & = 36\checkmark \end{align*}\]

Therefore, the solution set is \(\{e^9\}\).

Example 86

Solve \(\log_3(x^3-5)=1\).

Solution:

We will compose both sides by the exponential function base \(3\). Then solve for \(x\).

\[\begin{align*} 3^{\log_3(x^3-5)} & = 3^1\\ x^3-5 & = 3\\ x^3 & = 8\\ x & = 2 \end{align*}\]

Next, we will check the solution.

\[\begin{align*} \log_3((2)^3-5) & = 1\\ \log_3(8-5) & = 1\\ \log_3(3) & = 1\checkmark \end{align*}\]

Therefore, the solution set is \(\{2\}\).

Example 87

Solve \(\log(2x+1)+\log(x) = \log(x+8)\).

First, we will simplify the left-hand side of the equation.

\[\begin{align*} \log(2x+1)+\log(x) & = \log(x(2x+1))\\ & = \log(2x^2+x) \end{align*}\]

The equation to solve is now: \(\log(2x^2+x) = \log(x+8)\). Compose both sides of the equation by exponential function base \(10\) since \(10^{\log(x)}=x\).

\[\begin{align*} 10^{\log(2x^2+x)} & = 10^{\log(x+8)}\\ 2x^2 + x & = x + 8\\ 2x^2 - 8 & = 0\\ 2(x^2-4) & = 0\\ 2(x-2)(x+2) & = 0 \end{align*}\]

The solutions to \(2(x-2)(x+2)=0\) are \(x=2\) and \(x=-2\). However, the original equation is undefined when \(x=-2\) since \(\log(-2)\) is undefined. Next, we will check the solution \(x=2\).

\[\begin{align*} \log(2(2)+1)+\log(2) & = \log(2+8)\\ \log(5)+\log(2) & = \log(10)\\ \log(5\cdot 2) & = 1\\ \log(10) & = 1\checkmark \end{align*}\]

Therefore, the solution set is \(\{2\}\).

Remember to always check you solutions for log equations.