Lesson 14

To solve the equation \(5 \boldcdot e^{3a} = 90\), Lin wrote the following:

\(\displaystyle \begin {align}5 \boldcdot e^{3a} &= 90\\ e^{3a} &= 18\\ 3a &= \log_e 18\\ a&=\frac {\log_e 18}{3}\\ \end {align}\)

Is her solution valid? Be prepared to explain what she did in each step to support your answer.

1. Complete the table with equivalent equations. The first row is completed for you.

   	exponential form	logarithmic form
   a.	\(e^0= 1\)	\(\ln 1=0\)
   b.	\(e^1= e\)
   c.	\(e^\text{-1} = \frac{1}{e}\)
   d.		\(\ln \frac{1}{e^2} = \text-2\)
   e.	\(e^x = 10\)

2. Solve each equation by expressing the solution using \(\ln\) notation. Then, find the approximate value of the solution using the “ln” button on a calculator.
   1. \(e^m=20\)
   2. \(e^n=30\)
   3. \(e^p=7.5\)

Without using a calculator, solve each equation. It is expected that some solutions will be expressed using log notation. Be prepared to explain your reasoning.

1. \(10^x = 10,\!000\)
2. \(5 \boldcdot 10^x = 500\)
3. \(10^{(x+3)} = 10,\!000\)
4. \(10^{2x} = 10,\!000\)
5. \(10^x = 315\)
6. \(2 \boldcdot 10^x = 800\)
7. \(10^{(1.2x)} = 4,\!000\)
8. \(7\boldcdot 10^{(0.5x)} = 70\)
9. \(2 \boldcdot e^x=16\)
10. \(10 \boldcdot e^{3x}=250\)

So far we have solved exponential equations by

• finding whole number powers of the base (for example, the solution of \(10^x = 100,\!000\) is 5)
• estimation (for example, the solution of \(10^x = 300\) is between 2 and 3)
• using a logarithm and approximating its value on a calculator (for example, the solution of \(10^x = 300\) is \(\log 300 \approx 2.48\))

Sometimes solving exponential equations takes additional reasoning. Here are a couple of examples.

In the first example, the power of 10 is multiplied by 5, so to find the value of \(x\) that makes this equation true each side was divided by 5. From there, the equation was rewritten as a logarithm, giving an exact value for \(x\).

In the second example, the expressions on each side of the equation were rewritten as powers of 10: \(10^{(0.2t)}=10^3\). This means that the exponent \(0.2t\) on one side and the 3 on the other side must be equal, and leads to a simpler expression to solve where we don't need to use a logarithm.

How do we solve an exponential equation with base \(e\), such as \(e^x = 5\)? We can express the solution using the natural logarithm, the logarithm for base \(e\). Natural logarithm is written as \(\ln\), or sometimes as \(\log_e\). Just like the equation \(10^2 =100 \) can be rewritten, in logarithmic form, as \(\log_{10}100 = 2\), the equation \(e^0 = 1\) and be rewritten as \(\ln 1 = 0\). Similarly, \(e^{\text-2} = \frac{1}{e^2}\) can be rewritten as \(\ln \frac{1}{e^2} = \text{-}2\).

All this means that we can solve \(e^x = 5\) by rewriting the equation as \(x = \ln 5\). This says that \(x\) is the exponent to which base \(e\) is raised to equal 5.

To estimate the size of \(\ln 5\), remember that \(e\) is about 2.7. Because 5 is greater than \(e^1\), this means that \(\ln 5\) is greater than 1. \(e^2\) is about \((2.7)^2\) or 7.3. Because 5 is less than \(e^2\), this means that \(\ln 5\) is less than 2. This suggests that \(\ln 5\) is between 1 and 2. Using a calculator we can check that \(\ln 5 \approx 1.61\).