We need to solve the equation

[latex]400 = 245(1.03)^n[/latex]                        Begin by dividing both sides by [latex]245[/latex] to isolate the exponential

[latex]1.633 = 1.03^n[/latex]                                Now take the log of both sides

[latex]log(1.633) = log(1.03^n)[/latex]            Use the exponent property of logs on the right side

[latex]log(1.633)= n log(1.03)[/latex]           Now we can divide by [latex]log(1.03)[/latex]

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

[latex]n ≈ 16.591[/latex]

A full walk-through of this problem is available here.

You can view the transcript for “Using logs to solve for time” here (opens in new window).

In this problem, our “population” is the number of particles of pollutant per gallon. The initial pollutant is 10 million particles per gallon, so [latex]P_0 = 10,000,000[/latex]. Instead of changing with time, the pollutant changes with the number of filters, so [latex]n[/latex] 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 [latex]r = -0.90[/latex].We can then write the explicit equation for the pollutant:

[latex]P_n = 10,000,000(1 – 0.90)^{n} = 10,000,000(0.10)^{n}[/latex]

To solve the question of how many filters are needed to lower the pollutant to [latex]500[/latex] particles per gallon, we can set [latex]P_n[/latex] equal to [latex]500[/latex], and solve for [latex]n[/latex].

[latex]500 = 10,000,000(0.10)n[/latex]                     Divide both sides by [latex]10,000,000[/latex]

[latex]0.00005 = 0.10n[/latex]                                     Take the log of both sides

[latex]log(0.00005) = log(0.10n)[/latex]                    Use the exponent property of logs on the right side

[latex]log(0.00005) = n log(0.10)[/latex]                   Evaluate the logarithms to a decimal approximation

[latex]-4.301 = n (-1)[/latex]                                          Divide by [latex]-1[/latex], the value multiplying [latex]n[/latex]

[latex]4.301 = n[/latex]

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

More details about solving this scenario are available in this video.

You can view the transcript for “Using logs to solve an exponential” here (opens in new window).