In prior lessons, you used an exponential model to predict the population of a town based on a constant growth rate such as 6% per year.
In the real world however, populations often do not just grow continuously and without limit. A town originally founded near a convenient water source may grow very quickly at first, but the expansion will slow dramatically as houses and businesses run out of room near the water source, and need to begin transporting water further and further away.
How can a situation like this be modeled with an equation?
Watch This
Embedded Video:
 James Sousa: Ex: Exponential Function Application with Logarithms
Guidance
In a prior lesson, we considered the solutions of simple log equations. Now we return to that topic and explore some more complex examples. Solving more complicated log equations can be less difficult than you might think, by using our knowledge of log properties (for a review of the properties of logs, see the lesson: "Expanding and Condensing Logarithms").
For example, consider the equation log_{2} (x) + log_{2} (x  2) = 3. We can solve this equation using a log property.
log_{2} (x) + log_{2} (x  2) = 3  

log_{2} (x(x  2)) = 3  log_{b} x + log_{b} y = log_{b}(xy) 
log_{2} (x^{2}  2x) = 3 \begin{align*}\Rightarrow\end{align*} 
write the equation in exponential form. 
2^{3} = x^{2}  2x  
x^{2}  2x  8 = 0  Solve the resulting quadratic 
(x  4) (x + 2) = 0  
x = 2, 4 
The resulting quadratic has two solutions. However, only x = 4 is a solution to our original equataion, as log_{2}(2) is undefined. We refer to x = 2 as an extraneous solution.
Example A
Solve each equation
a. log (x + 2) + log 3 = 2  b. ln (x + 2)  ln (x) = 1 

Solution:
a. log (x + 2) + log 3 = 2

log (3(x + 2)) = 2 log_{b} x + log_{b} y = log_{b} (xy) log (3x + 6) = 2 Simplify the expression 3(x+2) 10^{2} = 3x + 6 Write the log expression in exponential form 100 = 3x + 6 3x = 94 Solve the linear equation x = 94/3
b. ln (x + 2)  ln (x) = 1

\begin{align*}ln \left (\frac{x + 2} {x}\right ) = 1\end{align*} ln(x+2x)=1 \begin{align*}log_b x  log_b y = log_b \left (\frac{x} {y}\right )\end{align*} logbx−logby=logb(xy) \begin{align*}e^1 = \frac{x + 2} {x}\end{align*} e1=x+2x Write the log expression in exponential form. \begin{align*}ex = x + 2\end{align*} ex=x+2 Multiply both sides by \begin{align*}x\end{align*} x .\begin{align*}ex  x = 2\end{align*} ex−x=2 Factor out \begin{align*}x\end{align*} x .\begin{align*}x (e  1) = 2\end{align*} x(e−1)=2 Isolate \begin{align*}x\end{align*} x .\begin{align*}x = \frac{2} {e  1}\end{align*} x=2e−1
The solution above is an exact solution. If we want a decimal approximation, we can use a calculator to find that x ≈ 1.16. We can also use a graphing calculator to find an approximate solution. Consider again the equation ln (x + 2)  ln (x) = 1. We can solve this equation by solving a system:
\begin{align*}\begin{cases}
y = ln (x + 2)  ln (x)\\
y = 1\\
\end{cases} \end{align*}
If you graph the system on your graphing calculator, you should see that the curve and the horizontal line intersection at one point. Using the INTERSECT function on the CALC menu (press <TI font_2nd>[CALC]), you should find that the x coordinate of the intersection point is approximately 1.16. This method will allow you to find approximate solutions for more complicated log equations.
Example B
Use a graphing calculator to solve each equation:
a. log(5  x) + 1 = log x  b. log_{2}(3x + 8) + 1 = log_{3} (10  x) 

Solution:
a. log(5  x) + 1 = log x
 The graphs of y = log (5  x) + 1 and y = log x intersect at x ≈ 4.5454545
 Therefore the solution of the equation is x ≈ 4.54.
b. log_{2} (3x + 8) + 1 = log_{3} (10  x)

First, in order to graph the equations, you must rewrite them in terms of a common log or a natural log. The resulting equations are: \begin{align*}y = \frac{log(3x + 8)}{log 2} + 1 \end{align*}
y=log(3x+8)log2+1 and \begin{align*}y = \frac{log(10  x)}{log 3}\end{align*}y=log(10−x)log3 . The graphs of these functions intersect at x ≈ 1.87. This value is the approximate solution to the equation.
Example C
Consider population growth:
Table 2
Year  Population 

1  2000 
5  4200 
10  6500 
20  8800 
30  10500 
40  12500 
If we plot this data, we see that the growth is not quite linear, and it is not exponential either.
Solution:
We can find a logarithmic function to model this data. First enter the data in the table in L1 and L2. Then press STAT to get to the CALC menu. This time choose option 9. You should get the function y = 930.4954615 + 2780.218173 ln x. If you view the graph and the data points together, as described in the Technology Note above, you will see that the graph of the function does not touch the data points, but models the general trend of the data.
Note about technology: you can also do this using an Excel spreadsheet. Enter the data in a worksheet, and create a scatterplot by inserting a chart. After you create the chart, from the chart menu, choose “add trendline.” You will then be able to choose the type of function. Note that if you want to use a logarithmic function, the domain of your data set must be positive numbers. The chart menu will actually not allow you to choose a logarithmic trendline if your data include zero or negative x values. See below:
Concept question wrapup A population which increases continuously at a constant rate may be modeled with an exponential function. A population which increases rapidly and then levels off may be modeled with a logarithmic function. 

Vocabulary
An extraneous solution is a solution to a function that is not useful in the given context.
Guided Practice
1) Solve for \begin{align*}x\end{align*}
2) Solve for \begin{align*}x\end{align*}
3) Solve for \begin{align*}x\end{align*}
4) Biologists use the formula \begin{align*}n = k \cdot log A\end{align*}
Answers
1) To solve \begin{align*}log_3 x + log_3 (x + 3) = 9\end{align*}

\begin{align*}log_3 (x (x + 3)) = 9\end{align*}
log3(x(x+3))=9 : Using \begin{align*}log_x y + log_x z = log_x (yz)\end{align*}logxy+logxz=logx(yz) 
\begin{align*}log_3 (x^2+3x) = 9\end{align*}
log3(x2+3x)=9 : Distribute the \begin{align*}x\end{align*}x 
\begin{align*}3^9 = x^2 + 3x\end{align*}
39=x2+3x : Write in exponential form 
\begin{align*}19,683 = x^2 + 3x\end{align*}
19,683=x2+3x : With a calculator 
\begin{align*}0 = x^2 + 3x  19,683\end{align*}
0=x2+3x−19,683 : Set = 0 
\begin{align*}x = 3 \pm 280.6\end{align*}
x=−3±280.6 : By the Quadratic Formula
\begin{align*}\therefore x = 277.6\end{align*} Since there can be no number raised to a power that would equal \begin{align*}283.6\end{align*}
2) To solve \begin{align*}log_2 x  log_2 (x  4) = 12\end{align*}
 \begin{align*}log_2 \frac{x}{x  4} = 12\end{align*} : Using \begin{align*}log_x y  log_x z = log_x \frac{y}{z}\end{align*}
 \begin{align*}2^{12} = \frac{x}{x  4}\end{align*} : Write in exponential form
 \begin{align*}4,096 = \frac{x}{x  4}\end{align*} : With a calculator
 \begin{align*}4,096x  16384 = x\end{align*} : Multiply both sides by \begin{align*}x  4\end{align*}
 \begin{align*}4,095x = 16,384\end{align*} : Simplify
 \begin{align*}x = 4\end{align*} : Divide
3) To solve \begin{align*}log (x  4) + log 6 = 4\end{align*}
 \begin{align*}log (x  4) \cdot 6 = 4\end{align*}
 \begin{align*}log (6x  24) = 4\end{align*}
 \begin{align*}10^4 = 6x  24\end{align*}
 \begin{align*}1000 = 6x 24\end{align*}
 \begin{align*}976 = 6x\end{align*}
 \begin{align*}162.67\end{align*}
4) To find the number of species in an area of 950km^{2}:
 \begin{align*}n = 943 \cdot log 950\end{align*} : Substitute the given \begin{align*}k\end{align*} and A values
 \begin{align*}n = 943 \cdot 2.977\end{align*} : With a calculator
 \begin{align*}n = 2,807\end{align*}
Therefore 2,807 species would likely live in the area.
Practice
Express problems 17 in exponential form:
 \begin{align*}log_{12} \frac{1}{1728} = 3\end{align*}
 \begin{align*}log_{216} 6 = \frac{1}{3}\end{align*}
 \begin{align*}log_{\frac{1}{3}} \frac{1}{9} = 3\end{align*}
 \begin{align*}log_{\frac{1}{4}} \frac{1}{16} = 2\end{align*}
 \begin{align*}log_5 125 = 3\end{align*}
 \begin{align*}log_{15} 225 = 2\end{align*}
 \begin{align*}log_{25} 5 = \frac{1}{2}\end{align*}
For problems 813, solve for \begin{align*}x\end{align*}
 \begin{align*}log_x 64 = 2\end{align*}
 \begin{align*}log_3 6561 = x\end{align*}
 \begin{align*}log_5 x = 4\end{align*}
 \begin{align*}log_x 27 = 3\end{align*}
 \begin{align*}log_2 x = 6\end{align*}
 \begin{align*}log_4 64 = x\end{align*}
For problems 1419, solve for \begin{align*}x\end{align*}
 \begin{align*}4 log (\frac{x}{5}) + log (\frac{625}{4}) = 2 log x\end{align*}
 \begin{align*}log_5 z + \frac{log_5 125}{log_5 x} = \frac{7}{2}\end{align*}
 \begin{align*}log p = \frac{2  log p}{log p}\end{align*}
 \begin{align*}2 log x  2 log (x+1) = 0\end{align*}
 \begin{align*}log (25  z^3)  3log (4  z) = 0\end{align*}
 \begin{align*}\frac{log (35  y^3)}{log (5  y)} = 3\end{align*}