Title: Default for 6.4 The Logarithmic Function
Course: HELM Workbooks 2023-2024

Exercise

Solve the equation \([value]=[value]\). To write your answer as a natural logarithm please write \(\log(n)\).

\( x= \)
Feedback: We have that \([value]=[value]\) implies \(e^x = \sqrt[value]\) where we take the positive square root as exponential functions are always positive. Taking logarithms of both sides gives \[x = \ln(\sqrt[value]) = \frac{\ln([value])}{2}.\]

The logarithmic function

Recall the operation of taking logarithms which reverses the operation of exponentiation. \[\qquad \text{If}\ a>0 \ \text{and}\ a\ne 1\ \ \text{then}\qquad x=a^y\qquad\text{implies}\qquad y=\log_ax\] In this quiz we consider the log function in more detail. We shall concentrate only on the functions \(\log x\) (i.e. to base \(10\)) and \(\ln x\) (i.e. to base \(e\)). The functions \(y=\log x\) and \(y=\ln x\) have similar characteristics. We can never choose \(x\) as a negative number since \(10^y\) and \(e^y\) are each always positive. The graphs of \(y=\log x\) and \(y=\ln x\) are shown in the figure below.

[Image: logarithmic and exponential functions figure]

Logarithmic and exponential functions

From the graphs we see that both functions are one-to-one so each has an inverse - the inverse function of \(\log_ax\) is \(a^x\). Let us do this for logs to base \(10\).

Exercise

Solve the equation \([value]=[value]e^x\)

\( x= \)
Feedback: We have that \[[value]=[value]e^x \text{ implies } \frac{[value]}{e^x} = [value].\] Hence \[\frac{[value]}{e^x} = e^{2x-x} = e^x = [value]\] and so, by taking logarithms of both sides \[x = \ln([value]).\]

Example

Find \(x\) if \(10^x-5+6(10^{-x})=0\).

Solution

We first simplify this expression by multiplying through by \(10^x\) (to eliminate the term \(10^{-x}\)): \[10^x(10^x)-10^x(5)+10^x(6(10^{-x}))=0,\] or \[(10^{x})^2-5(10^x)+6=0\qquad\text{since}\qquad 10^x(10^{-x})=10^0=1.\]

We realise that this expression is a quadratic equation. Let us put \(y=10^x\) to give \[y^2-5y+6=0.\]

Now, we can factorise to give \[(y-3)(y-2)=0 \qquad \text{so that}\qquad y=3\quad\text{or}\quad y=2.\]

For each of these values of \(y\) we obtain a separate value for \(x\) since \(y=10^x\).

Case 1 If \(y=3\) then \(3=10^x\) implying \(x=\log_{10} 3=0.4771212\).

Case 2 If \(y=2\) then \(2=10^x\) implying \(x=\log_{10} 2=0.3010300\).

We conclude that the equation \(10^x-5+6(10^{-x})=0\) has two possible solutions for \(x\): either \(x=\log_{10} 3\) or \(x=\log_{10} 2\).

Exercise

Solve the following equation for \(x\).

\[\displaystyle{[value]=0}\]

\(x=\)

\(x=\)
Feedback: First write this equation as a quadratic in the variable \(y=e^x\) remembering that \(e^{2x}\equiv (e^x)^2\): \[[value] \, \text{ becomes } \, [value]=0.\] Now solve the quadratic for \(y\): \[[value]=0 \, \text{ implies } \, (y-[value])(y-[value])=0.\] Therefore \(y=[value]\) or \(y=[value].\) Finally, for each of your values of \(y\), find \(x\): Case 1 If \(y=[value]\) then \([value]=e^x\) implies \(x=\ln [value]\); Case 2 If \(y=[value]\) then \([value]=e^x\) implies \(x=\ln [value].\)

Exercise

The temperature \(T\), in degrees C, of a chemical reaction is given by the formula \[\displaystyle{T=[value]e^{[value]t}}\] where \(t\geq 0\) is the time, in seconds.

Calculate the time taken for the temperature to reach \([value]^\circ \text{C}.\) Please give your answer to 3 significant figures.

\( t = \) \( \textrm{s}\)
Feedback: We solve \[[value]=[value]e^{[value]t}.\] Rearranging gives \[e^{[value]t} = \frac{[value]}{[value]}\] and taking logarithms \[t = \frac{1}{[value]} \ln\left(\frac{[value]}{[value]}\right) \approx [value] .\] So the time is [value] seconds (to 3 s.f.).

Arrhenius' law

Introduction

Chemical reactions are very sensitive to temperature; normally, the rate of reaction increases as temperature increases. For example, the corrosion of iron and the spoiling of food are more rapid at higher temperatures. Chemically, the probability of collision between two molecules increases with temperature, and an increased collision rate results in higher kinetic energy, thus increasing the proportion of molecules that have the activation energy for the reaction, i.e. the minimum energy required for a reaction to occur. Based upon his observations, the Swedish chemist, Svante Arrhenius, proposed that the rate of a chemical reaction increases exponentially with temperature. This relationship, now known as Arrhenius' law, is written as

\[k = k_0 \, \exp \left( \frac{-E_a}{R\,T} \right) \qquad (1)\]

where \(k\) is the reaction rate constant, \(k_0\) is the frequency factor, \(E_a\)is the activation energy, \(R\) is the universal gas constant and \(T\) is the absolute temperature. Thus, the reaction rate constant, \(k\), depends on the quantities \(k_0\) and \(E_a\), which characterise a given reaction, and are generally assumed to be temperature independent.

Problem in words

In a laboratory, ethyl acetate is reacted with sodium hydroxide to investigate the reaction kinetics. Calculate the frequency factor and activation energy of the reaction from Arrhenius' Law, using the experimental measurements of temperature and reaction rate constant in the table:

\(T\) | 310 | 350
\(k\) | 7.757192 | 110.9601

Mathematical statement of problem

Given that \(k = 7.757192 s^{-1}\) at \(T = 310 K\) and \(k = 110.9601 s^{-1}\) at \(T = 350 K\), use Equation (1) to produce two linear equations in \(E_a\) and \(k_0\). Solve these to find \(E_a\) and \(k_0\). (Assume that the gas constant \(R = 8.314 J K^{-1} \text{mol}^{-1}.\))

Mathematical analysis

Taking the natural logarithm of both sides of (1)

\[\ln k = \ln \left\{k_0 \, \exp\left(\frac{-E_a}{R\,T}\right)\right\}= \ln k_0 - \frac{E_a}{R\,T}.\] Now inserting the experimental data gives the two linear equations in \(E_a\) and \(k_0\) \[\ln k_1 = \ln k_0 - {\frac{E_a}{R\,T_1}} \qquad (2)\] \[\ln k_2 = \ln k_0 - \frac{E_a}{R\,T_2} \qquad (3)\] where \(k_1 = 7.757192\), \(T_1 = 310\) and \(k_2 = 110.9601\), \(T_2 = 350\).

Firstly, to find \(E_a\), subtract Equation (2) from Equation (3) \[\ln k_2-\ln k_1= \dfrac{E_a}{R\,T_1}-\dfrac{E_a}{R\,T_2}= \dfrac{E_a}{R} \left( \dfrac{1}{T_1} - \dfrac{1}{T_2} \right)\] so that \[E_a = \frac{R\,(\ln k_2-\ln k_1)}{\left(\frac{1}{T_1}-\frac{1}{T_2}\right)}\] and substituting the values gives \[E_a= 60 000 \mbox{ J mol}^{-1} = 60 \mbox{ kJ mol}^{-1}\] Secondly, to find \(k_0\), from (2) \[\ln k_0 = \ln k_1 + \frac{E_a}{R\,T_1}\] which implies \[k_0 = \exp \left( \ln k_1 + \frac{E_a}{R\,T_1}\right)= k_1 \, \exp \left(\frac{E_a}{R\,T_1}\right)\] and substituting the values gives \[k_0= 1.0 \times 10^{11} \mbox{ s}^{-1}.\]

Exercise

For a chemical reaction with frequency factor \(k_0 = 0.5 \, \text{s}^{-1}\) and ratio \(E_a/R=[value] \, \text{K}\), use Arrhenius' law to find the temperature at which the reaction rate constant would be equal to \(0.1 \, \text{s}^{-1}\). Please give your answer to \(3\) s.f.

\( T = \) \( \text{K}\)
Feedback: Rearranging Equation (1) \[\frac{k}{k_0} = \exp \left( \frac{-E_a}{R\,T} \right).\] Taking the natural logarithm of both sides \[\ln \left( \frac{k}{k_0} \right) = \frac{-E_a}{R\,T}\] so that \[T = \frac{-E_a}{R\,\ln \left( k/k_0 \right)}= \frac{E_a}{R\,\ln \left( k_0/k \right)}.\] Substituting the values gives \(T=[value] \mbox{ K}.\)

Exercise

Find the solution for \(x\) if \(\tanh x=\displaystyle[value]\). Give a precise answer.

\( x = \)
Feedback: First re-write \(\tanh x\) in terms of exponentials: \[\tanh x=\dfrac{e^x-e^{-x}}{e^x+e^{-x}}=\dfrac{e^{2x}-1}{e^{2x}+1}\] Now substitute into \(\tanh x=[value]\): \(\tanh x=[value]\) implies \[\dfrac{e^{2x}-1}{e^{2x}+1}=[value]\] so, on simplifying, \[\dfrac{e^{2x}-1}{e^{2x}+1}=[value] \text{ implies } (e^{2x}-1)=\dfrac{1}{[value]}(e^{2x}+1)\] so \(\dfrac{([value]-1)e^{2x}}{[value]}=\dfrac{[value]+1}{[value]}\) so, finally, \(e^{2x}=\frac{[value]+1}{[value]-1}\). Now complete your solution by finding \(x\): \[e^{2x}=\frac{[value]+1}{[value]-1} \text{ so } x= \dfrac{1}{2}\ln \left(\frac{[value]+1}{[value]-1}\right) = \dfrac{1}{2}\ln \left(\frac{[value]}{[value]}\right).\] Alternatively, many calculators can directly calculate the inverse function \(\tanh^{-1}\). If you have such a calculator then you can use the fact that \[\tanh x=[value] \qquad \text{implies} \qquad x=\tanh^{-1}[value]\] to obtain \(x\) directly.

Exercise

Solve for the variable \(x\). To write your answer as a logarithm to base \(n\), please write log_n(m).

1. \([value] =10^x\)
2. \(10^{-x/[value]}=[value]\)
3. \(\dfrac{1}{[value]-\pi^x}=[value]\). Please do not use logarithms to base \(\pi\) in your answer.
Feedback: 1. \([value] =10^x\) We take logs of base \(10\) on both sides: \(\log(10^x) =\log[value]\pi \Rightarrow\) \(x \log(10) =\log[value]\pi \Rightarrow\) \(x =\log[value]\pi.\) 2. \(10^{-x/[value]}=[value]\) We take logs of base \(10\) on both sides: \(\log(10^{-x/[value]}) = \log[value] \Rightarrow\) \( -\dfrac{x}[value] \log10 = \log[value] \Rightarrow\) \(x=-[value]\log [value].\) 3. \(\dfrac{1}{[value]-\pi^x}=[value]\) Here we simplify the expression before taking logs of base \(10\) on both sides: \([value]([value]-\pi^x)=1 \Rightarrow\) \(\pi^x=[value] \Rightarrow\) \(\log(\pi^x)= \log[value] \Rightarrow\) \(x \log(\pi)= \log[value] \Rightarrow\) \(x=[value].\)

Exercise

The reaction

2NO\(_2\)(g) \(\longrightarrow\) 2NO(g) + O\(_2\)(g)

has a reaction rate constant of \(1.0 \times 10 ^{-10}\) \(\text{s}^{-1}\) at \(300 \text{K}\) and activation energy of \(111 \text{kJ mol}^{-1} = 111 000\text{ J mol}^{-1}\). Use Arrhenius' law to find the reaction rate constant at a temperature of [value]\(\text{K}\). Please give your answer to \(4\) s.f.

\( k = \)
Feedback: Rearranging Arrhenius' equation gives \[k_0 = k \, \exp\left(\frac{E_a}{R\,T}\right).\] Substituting the values gives \[k_0= 2.126 \times 10^{9} \mbox{ s}^{-1}.\] Now we use this value of \(k_0\) with \(E_a\) in Arrhenius' equation (1) to find \(k\) at \(T\) = [value] K \[k= k_0 \, \exp \left( \frac{-E_a}{R\,T} \right)= [value] \mbox{ s}^{-1}.\]

Exercise

Solve the equations

1. \(e^{2x}=[value]e^x\)
2. \([value]=0\)
3. \(\cosh x=[value]\)
\(x=\)
\(x=\)
Feedback: 1. \(e^{2x}=[value]e^x\) Take logs of base \(e\) on both sides: \(\ln(e^{2x}) = \ln([value]e^x) \Rightarrow\) \(2x=\ln [value]+x \Rightarrow x=\ln [value].\) 2. \([value]=0\) Let \(y=e^x,\) then the equation becomes \([value]=0.\) By solving the quadratic we find: \(y=[value]\) or \(y=[value]\) (we cannot take the negative root since exponentials can never be negative). Thus \(e^x= [value] \Rightarrow x=\ln ([value]).\) 3. \(e^x+e^{-x}=[value]\)can be written as \([value]=0.\) Let \(y=e^x,\) then the equation becomes \([value] =0. \) By solving the quadratic we find: \(y = e^x = \dfrac{[value] \pm \sqrt{([value])^2-4}}{2} = [value] \pm \sqrt{[value]^2-1}.\) We accept both roots since both are positive. Finally we have \(x=[value]\) or \(x=[value].\)

Solving equations involving logarithms and exponentials

To solve equations which involve logarithms or exponentials we need to be aware of the basic laws which govern both of these mathematical concepts. We illustrate by considering some examples.

Example

Solve for the variable \(x\).

1. \(3=10^x\)
2. \(10^{x/4}=\log_{10} 3\)
3. \(\dfrac{1}{17-e^x}=4\)

Solution

1. Here we take logs (to base \(10\) because of the term \(10^x\)) of both sides to get \[ \log 3=\log 10^x=x\log 10=x \] where we have used the general property that \(\log_aA^k=k\log_aA\) and the specific property that \(\log 10=1\). Hence \(x=\log 3\) or, in numerical form, \(x=0.47712\) to \(5\) d.p.
2. The approach used in (a) is used here. Take logs of both sides: \[\log_{10}(10^{x/4})=\log_{10}(\log_{10}3)\] that is \[\dfrac{x}{4}\log_{10} 10=\log_{10} (\log_{10} 3)=\log_{10} (0.4771212)=-0.3213712.\] So, since \(\log_{10} 10=1\), we have \(x=4(-0.3213712)=-1.28549\) to \(5\) d.p.
3. Here we simplify the expression before taking logs. \[\dfrac{1}{17-e^x}=4\qquad\text{implies}\qquad 1=4(17-e^x)\] or \(4e^x=4(17)-1=67\) so \(e^x=16.75\). Now taking natural logs of both sides (because of the presence of the \(e^x\) term) we have: \[\ln (e^x)=\ln (16.75)=2.8183983.\] But \(\ln (e^x)=x\ln e=x\) and so the solution to \(\dfrac{1}{17-e^x}=4\) is \(x=2.81840\) to \(5\) d.p.

Solving equations involving hyperbolic functions

As a final example we consider equations involving the hyperbolic functions.

Example

Solve the equations :

1. \(\sinh 3x=0\)
2. \(\cosh 3x=2\)
3. \(2\cosh^2x=3\cosh 2x-3\)

Solution

1. From its graph we know that \(\sinh x=0\) only when \(x=0\), so we need \(3x=0\) which implies \(x=0\).
2. \[\cosh 3x=2 \qquad \text{ implies } \qquad \dfrac{e^{3x}+e^{-3x}}{2}=2 \qquad \text{ or } \qquad e^{3x}+e^{-3x}-4=0.\] Now multiply through by \(e^{3x}\) (to eliminate the term \(e^{-3x}\)) to give \[e^{3x}e^{3x}+e^{3x}e^{-3x}-4e^{3x}=0 \qquad \text{ or } \qquad (e^{3x})^2-4e^{3x}+1=0.\] This is a quadratic equation in the variable \(e^{3x}\) so substituting \(y=e^{3x}\) gives \[y^2-4y+1=0\qquad \text{implying}\qquad y=2\pm\sqrt{3}\quad \text{so}\quad y=3.7321\quad\text{or}\quad 0.26795\] \[ \qquad e^{3x}=3.7321\qquad \text{implies}\qquad x=\dfrac{1}{3}\ln 3.7321=0.439\quad\text{to 3 d.p.}\] \[ \qquad e^{3x}=0.26795\qquad \text{implies}\qquad x=\dfrac{1}{3}\ln 0.26795=-0.439\quad\text{to 3 d.p.}\]
3. We first simplify this expression by using the identity: \(\cosh 2x=2\cosh^2-1\). Thus the original equation \(2\cosh^2x=3\cosh 2x-3\) becomes \(\cosh 2x+1=3\cosh 2x-3\) or, when written in terms of exponentials: \[\dfrac{e^{2x}+e^{-2x}}{2}=3\left(\dfrac{e^{2x}+e^{-2x}}{2}\right)-4.\] Multiplying through by \(2e^{2x}\) gives \(e^{4x}+1=3(e^{4x}+1)-8e^{2x}\) or, after simplifying: \[e^{4x}-4e^{2x}+1=0.\] Writing \(y=e^{2x}\) we easily obtain \(y^2-4y+1=0\) with solution (using the quadratic formula): \[y=\dfrac{4\pm \sqrt{16-4}}{2}=2\pm\sqrt{3}\] If \(y=2+\sqrt{3}\) then \(2+\sqrt{3}=e^{2x}\) implying \(x=0.65848\) to \(5\) d.p. If \(y=2-\sqrt{3}\) then \(2-\sqrt{3}=e^{2x}\) implying \(x=-0.65848\) to \(5\) d.p.

Example

Solve for \(x\) if \(3\ln x+4\log_{10} x=1\).

Solution

This has logs to two different bases. So we must first express each logarithm in terms of logs to the same base, \(e\) say Recall

\[\log_{10} x=\dfrac{\ln x}{\ln 10}.\] So \(3\ln x+4\log x=1\) becomes \[ 3\ln x+4\dfrac{\ln x}{\ln 10}=1\qquad\text{or}\qquad \left(3+\dfrac{4}{\ln 10}\right)\ln x=1\] leading to \(\ln x=\dfrac{\ln 10}{3\ln 10 +4}=\dfrac{2.302585}{10.907755}=0.211096\) and so \[ x=e^{0.211096}=1.2350311.\]