### Unlimited growth: Exponential growth

### Solving exponential equations

Models of exponential growth often involve solving exponential equations. Therefore, we need logarithmic functions, both the natural logarithm and logarithms with a different base number (usually base 10). The following example of microbial growth illustrates this.

At time \(t=0\), a culture of bacteria consists of 100 bacteria and the number of bacteria increases every hour by a factor \(g=4.6\). The number \(A\) at time \(t\) is therefore modelled by the exponential function \[A(t)=100\cdot 4.6^t\] Determine at what time there are 10000 bacteria in the culture, according to this growth model. Give your answer in three decimals.

To determine at what time there are 10000 bacteria in the culture, according to this growth model, you need to solve \(A(t)=10000\), that is: \[100\cdot 4.6^t=10000\] This can be done in a couple of steps.

*Step 1.*Remove the product on the left by division: \[\begin{aligned}100\cdot 4.6^t=10000\;&\stackrel{\blue{\mathrm{\phantom{xx}divide\phantom{xx}}}}{\implies} \;4.6^t = \frac{10000}{100}\\ &\stackrel{\blue{\mathrm{\phantom{xx}expand\phantom{xx}}}}{\implies} 4.6^t=100\end{aligned}\] You could write down the solution straight away: \[t=\log_{4.6}(100),\] but this does not really help you because your calculator may not not have a button for a logarithm with base 4.6. The next two steps will help you further.*Step 2.*Apply the logarithm on both sides of the equationv and simplify. Because there is a power of 10 on the right side, the logarithm with base 10 is a convenient choice \[\begin{aligned}4.6^t=100\;&\stackrel{\blue{\phantom{xx}\log_{10}\phantom{xx}}}{\implies}\;\;\;\; \log_{10}(4.6^t) = \log_{10}(100) \\ &\stackrel{\blue{\mathrm{\phantom{xx}expand\phantom{xx}}}}{\implies} t \cdot \log_{10}(4.6) =2\end{aligned}\]*Step 3.*Now you only have to solve the linear equation. \[t \cdot \log_{10}(4.6) =2\stackrel{\blue{\mathrm{\phantom{xx}divide\phantom{xx}}}}{\implies} t=\frac{2}{\log_{10}(4.6)}\approx 3.018\;({}\approx3\;\mathrm{hour}\;1\;\mathrm{min})\]In the example above, we used the logarithm with base 10 to solve an exponential equation, because a power of 10 occurred in the formula and this reduced the computational effort. But we couls also have use the natural logarithm. The following example shows how.

A patient is given an intravenous bolus injection of \(500\;\mathrm{mg}\) antibiotic. This drug is spreading rapidly in the body. Suppose that at time \(t=0\), very shortly after the time of administration, the plasma concentration of antibiotic equals \(33.1\;\mathrm{mg/L}\). After this, the antibiotic is eliminated from the body through a process of exponential decay. Suppose the elimination rate constant per hour equals \(0.51\). This indicates an hourly growth factor of \(g=e^{-0.51}\approx 0.60\). The plasma concentration \(C\) (in mg/L) at time \(t\) (in hours) is, according to this model \[C(t)=33.1\cdot 0.60^t\] Determine when, according to this model, the plasma concentration is decreased to \(5\;\mathrm{mg/L}\). Give your answer in three decimals.

To determine at what time the plasma concentration has decreased to \(5\;\mathrm{mg/L}\) according to this exponential decay model, you should solve \(C(t)=5\): \[33.1\cdot 0.60^t=5\] This can be done in a couple of steps. For a change, we do something different than in the first example.

\[\begin{aligned} 33.1\cdot 0.60^t=5&\stackrel{\blue{\ln}}{\implies} \phantom{x}\ln(33.1\cdot 0.60^t) = \ln(5)\\ \\ &\stackrel{\mathrm{\blue{expand}}}{\implies} \ln(33.1)+t \cdot \ln(0.60) =\ln(5)

\end{aligned} \] This step shows that, by using a logarithm, you can reshape an exponential equation to a linear equation.

\[\begin{aligned} \ln(33.1)+t \cdot \ln(0.60) =\ln(5)\;&\stackrel{\blue{-\ln(33.1)}}{\implies} t\cdot \ln(0.60) =\ln(5)-\ln(33.1)\\ \\ &\stackrel{\blue{\div\ln(0.60)}}{\implies} t=\frac{\ln(5)-\ln(33.1)}{\ln(0.60)}\\ &\phantom{\stackrel{\blue{\div\ln(0.60)}}{\implies} t} \,\approx 3.700\;{}(\approx 3\;\mathrm{hours}\;42\;\mathrm{min})

\end{aligned}\] Delaying numerical calculations prevents rounding errors in intermediate results, which may eventually make a difference in the final answer. It also paves the way for a complete algebraic approach to a generic problem with parameters.

*Step 1.*Apply the natural logarithm to both sides and simplify.\[\begin{aligned} 33.1\cdot 0.60^t=5&\stackrel{\blue{\ln}}{\implies} \phantom{x}\ln(33.1\cdot 0.60^t) = \ln(5)\\ \\ &\stackrel{\mathrm{\blue{expand}}}{\implies} \ln(33.1)+t \cdot \ln(0.60) =\ln(5)

\end{aligned} \] This step shows that, by using a logarithm, you can reshape an exponential equation to a linear equation.

*Step 2.*Now all you have to do is solve this linear equation. To understand this process, let the logarithms intact for now:\[\begin{aligned} \ln(33.1)+t \cdot \ln(0.60) =\ln(5)\;&\stackrel{\blue{-\ln(33.1)}}{\implies} t\cdot \ln(0.60) =\ln(5)-\ln(33.1)\\ \\ &\stackrel{\blue{\div\ln(0.60)}}{\implies} t=\frac{\ln(5)-\ln(33.1)}{\ln(0.60)}\\ &\phantom{\stackrel{\blue{\div\ln(0.60)}}{\implies} t} \,\approx 3.700\;{}(\approx 3\;\mathrm{hours}\;42\;\mathrm{min})

\end{aligned}\] Delaying numerical calculations prevents rounding errors in intermediate results, which may eventually make a difference in the final answer. It also paves the way for a complete algebraic approach to a generic problem with parameters.

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