# Solve Exponentials for Time: Logarithms

Earlier, we found that since Olympia, WA had a population of 245 thousand in 2008 and had been growing at 3% per year, the population could be modeled by the equation
*Pn* = (1+0.03)*n *(245,000), or equivalently, *Pn* = 245,000(1.03)*n*.
Using this equation, we were able to predict the population in the future.
Suppose we wanted to know when the population of Olympia would reach 400 thousand. Since we are looking for the year *n* when the population will be 400 thousand, we would need to solve the equation
400,000 = 245,000(1.03)*n* dividing both sides by 245,000 gives
1.6327 = 1.03*n*
One approach to this problem would be to create a table of values, or to use technology to draw a graph to estimate the solution.
From the graph, we can estimate that the solution will be around 16 to 17 years after 2008 (2024 to 2025). This is pretty good, but we’d really like to have an algebraic tool to answer this question. To do that, we need to introduce a new function that will undo exponentials, similar to how a square root undoes a square. For exponentials, the function we need is called a logarithm. It is the inverse of the exponential, meaning it undoes the exponential. While there is a whole family of logarithms with different bases, we will focus on the common log, which is based on the exponential 10*x*.

*x*), undoes the exponential 10

*x*This means that log(10

*x*) =

*x*, and likewise 10log(

*x*) =

*x*This also means the statement 10

*a*=

*b*is equivalent to the statement log(

*b*) =

*a*log(

*x*) is read as “log of

*x*”, and means “the logarithm of the value

*x*”. It is important to note that this is

*not*multiplication – the log doesn’t mean anything by itself, just like √ doesn’t mean anything by itself; it has to be applied to a number.

### Example 9

*x*0 = 1. log(1) = log(100) = 0

*10 times bigger*for the log to increase in value by 1. Of course, most numbers cannot be written as a nice simple power of 10. For those numbers, we can evaluate the log using a scientific calculator with a log button.

### Example 10

### Example 11

*x*= 1000 b) Solve 10

*x*= 3 c) Solve 2(10

*x*) = 8 a) Taking the log of both sides gives log(10

*x*) = log(1000) Since the log undoes the exponential, log(10

*x*) =

*x*. Similarly log(1000) = log(103) = 3. The equation simplifies then to

*x*= 3. b) Taking the log of both sides gives log(10

*x*) = log(3). On the left side, log(10

*x*) =

*x*, so

*x*= log(3). We can approximate this value with a calculator.

*x*≈ 0.477 c) Here we would first want to isolate the exponential by dividing both sides of the equation by 2, giving 10

*x*= 4. Now we can take the log of both sides, giving log(10

*x*) = log(4), which simplifies to

*x*= log(4) ≈ 0.602

*n*from earlier, which have a base of 1.03? For that, we need the exponent property for logs.

### Example 12

### Example 13

*Pn*= 245(1.03)

*n*, where

*n*is years after 2008, and the population is measured in thousands. Find when the population will be 400 thousand. We need to solve the equation 400 = 245(1.03)

*n*Begin by dividing both sides by 245 to isolate the exponential 1.633 = 1.03

*n*Now take the log of both sides log(1.633) = log(1.03

*n*) Use the exponent property of logs on the right side log(1.633)=

*n*log(1.03) Now we can divide by log(1.03) [latex]\frac{\log(1.633)}{\log(1.03)}=n[/latex] We can approximate this value on a calculator

*n*≈ 16.591

### Example 14

*P0*= 10,000,000. Instead of changing with time, the pollutant changes with the number of filters, so

*n*will represent the number of filters the water passes through. Also, since the amount of pollutant is

*decreasing*with each filter instead of increasing, our “growth” rate will be negative, indicating that the population is decreasing instead of increasing, so

*r*= -0.90. We can then write the explicit equation for the pollutant:

*Pn*= 10,000,000(1 – 0.90)

*n*= 10,000,000(0.10)

*n*To solve the question of how many filters are needed to lower the pollutant to 500 particles per gallon, we can set

*Pn*equal to 500, and solve for

*n*. 500 = 10,000,000(0.10)

*n*Divide both sides by 10,000,000 0.00005 = 0.10

*n*Take the log of both sides log(0.00005) = log(0.10

*n*) Use the exponent property of logs on the right side log(0.00005) =

*n*log(0.10) Evaluate the logarithms to a decimal approximation -4.301 =

*n*(-1) Divide by -1, the value multiplying

*n*4.301 =

*n*It would take about 4.301 filters. Of course, since we probably can’t install 0.3 filters, we would need to use 5 filters to bring the pollutant below the desired level.

### Try it Now 4

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