UK A-Level: Exponentials and Logarithms

Exponential and logarithmic functions are not just abstract mathematical concepts; they are powerful tools that describe phenomena from radioactive decay and population growth to sound perception and financial interest. Mastering them is essential for A-Level mathematics, providing the foundation for calculus, advanced modeling, and numerous STEM applications.

Defining the Core Functions

An exponential function is any function of the form $y=a^x$, where the base $a$ is a positive constant ($a>0$) and $a\ne 1$. The key feature is that the variable $x$ is in the exponent. Its inverse function is the logarithm. The statement $y=a^x$ is mathematically equivalent to $x={\log }_{a}y$. We say "${\log }_{a}y$" as "the logarithm of $y$ to the base $a$". This means the logarithm answers the question: "To what power must we raise $a$ to get $y$?"

For example, since $2^3=8$, we know ${\log }_{2}8=3$. The most common bases are 10 (common logarithm, ${\log }_{10}$) and the special number $e\approx 2.71828$ (natural logarithm, $\ln$). The functions $y=e^x$ and $y=\ln x$ are a crucial inverse pair you will encounter constantly.

The Laws of Logarithms

Logarithms transform multiplicative relationships into additive ones, governed by three key laws. For any positive base $a$ ($a\ne 1$) and positive numbers $x$ and $y$:

1. The Addition Law: ${\log }_a(xy)={\log }_{a}x+{\log }_{a}y$.

The log of a product is the sum of the logs.

1. The Subtraction Law: ${\log }_a(\frac{x}{y})={\log }_{a}x-{\log }_{a}y$.

The log of a quotient is the difference of the logs.

1. The Power Law: ${\log }_a(x^k)=k{\log }_{a}x$.

The log of a power lets you bring the exponent down as a multiplier.

These laws are derived from the laws of indices and are your primary tools for manipulating and solving logarithmic equations. For instance, to combine ${\log }_{5}10+{\log }_{5}2$, you use the addition law: ${\log }_5(10\times 2)={\log }_{5}20$.

Solving Exponential and Logarithmic Equations

The strategy for solving equations where the variable is in an exponent is to use logarithms to "bring it down". Conversely, to solve equations with logarithms, you often rewrite them in exponential form.

Example: Solving an Exponential Equation
Solve $3^{2x-1}=50$.

1. Take the logarithm of both sides (any base, but $\ln$ is often easiest): $\ln (3^{2x-1})=\ln 50$.
2. Apply the power law: $(2x-1)\ln 3=\ln 50$.
3. Solve the linear equation: $2x-1=\frac{\ln 50}{\ln 3}$.
4. Therefore, $x=\frac{1}{2}\left(\frac{\ln 50}{\ln 3}+1\right)$.

Example: Solving a Logarithmic Equation
Solve ${\log }_4(x+3)+{\log }_4(2-x)=1$.

1. Use the addition law: ${\log }_4((x+3)(2-x))=1$.
2. Rewrite in exponential form: $(x+3)(2-x)=4^1$.
3. Expand and rearrange: $-x^2-x+6=4\Rightarrow -x^2-x+2=0$ or $x^2+x-2=0$.
4. Factorise: $(x+2)(x-1)=0$, so $x=-2$ or $x=1$.
5. Crucially, check against the original domain: ${\log }_{a}u$ is only defined for $u>0$.

For $x=-2$: $(x+3)=1>0$, but $(2-x)=4>0$. Valid. For $x=1$: $(x+3)=4>0$, but $(2-x)=1>0$. Valid. Both solutions are valid.

Natural Exponential and Logarithm Functions

The natural exponential function is $f(x)=e^x$, where $e$ is the irrational Euler's constant. It is unique because the gradient of $y=e^x$ at any point is equal to its value at that point. Its inverse is the natural logarithm, $g(x)=\ln x={\log }_{e}x$.

Their unique properties make them indispensable in calculus:

• $\frac{d}{dx}(e^x)=e^x$ and $\frac{d}{dx}(\ln x)=\frac{1}{x}$.
• They obey all standard logarithm laws: $\ln (ab)=\ln a+\ln b$, $\ln (a/b)=\ln a-\ln b$, $\ln (a^k)=k\ln a$.
• The key inverse identities are $e^{\ln x}=x$ (for $x>0$) and $\ln (e^x)=x$.

Modeling Growth and Decay

Many real-world processes are modeled by equations of the form $N=N_0{e}^{kt}$, where:

• $N$ is the final amount.
• $N_0$ is the initial amount (when $t=0$).
• $k$ is the rate constant.
• $t$ is time.

If $k>0$, the model describes exponential growth (e.g., unchecked bacterial growth). If $k<0$, it describes exponential decay (e.g., radioactive decay or cooling). The model implies a constant relative or percentage growth/decay rate. A classic application is finding the half-life $t_{1/2}$ of a decaying substance, the time for the quantity to halve. From $\frac{1}{2}{N}_0=N_0{e}^{k{t}_{1/2}}$, we derive $t_{1/2}=\frac{\ln 2}{-k}$.

Fitting Models Using Logarithmic Transformations

Given experimental data, how can you test if it follows an exponential relationship $y=a{b}^x$? You can use a logarithmic transformation to "linearise" the data.

1. Take the logarithm of both sides: $\log y=\log (a{b}^x)=\log a+x\log b$.
2. This is now in the form $Y=mX+c$, where $Y=\log y$, $m=\log b$, $X=x$, and $c=\log a$.
3. Plot $\log y$ against $x$. If the original relationship was exponential, this new graph will be a straight line.
4. From the gradient $m$ and y-intercept $c$ of this line, you can find the original constants: $b=10^m$ and $a=10^c$ (if using ${\log }_{10}$).

This technique is powerful for verifying models and estimating parameters from real data.

The Change of Base Formula

Calculators typically only have buttons for ${\log }_{10}$ and $\ln$. The change of base formula allows you to evaluate a logarithm with any base: $${\log }_{a}b=\frac{{\log }_{c}b}{{\log }_{c}a}$$ Here, $c$ can be any valid base you choose, usually 10 or $e$. This formula is also invaluable for solving equations with different bases.

Example: Evaluate ${\log }_{3}7$. Using the formula with base $e$: ${\log }_{3}7=\frac{\ln 7}{\ln 3}\approx \frac{1.94591}{1.09861}\approx 1.771$.

Example: Solve $2^{x+1}=5^{x-1}$.

1. Take $\ln$ of both sides: $\ln (2^{x+1})=\ln (5^{x-1})$.
2. Apply power law: $(x+1)\ln 2=(x-1)\ln 5$.
3. Expand: $x\ln 2+\ln 2=x\ln 5-\ln 5$.
4. Gather $x$ terms: $x\ln 2-x\ln 5=-\ln 5-\ln 2$.
5. Factor: $x(\ln 2-\ln 5)=-(\ln 5+\ln 2)$.
6. Solve: $x=\frac{-(\ln 5+\ln 2)}{(\ln 2-\ln 5)}=\frac{\ln (5\times 2)}{\ln (5/2)}=\frac{\ln 10}{\ln (2.5)}$.

Common Pitfalls

1. Misapplying Logarithm Laws: A common error is writing ${\log }_a(x+y)$ as ${\log }_{a}x+{\log }_{a}y$. This is false. The laws only apply to the log of a product or quotient. There is no simple law for the log of a sum.

1. Ignoring the Domain: The function ${\log }_{a}x$ is only defined for $x>0$. When solving logarithmic equations, any potential solution that leads to taking the log of a non-positive number must be rejected. Always check your solutions against the original equation's domain.

1. Incorrect Inverse Operations: Remember that $e^{\ln x}=x$ and $\ln (e^x)=x$. Confusing these, or trying to apply them to expressions with different bases, leads to errors. For example, $10^{\ln x}\ne x$ (unless in the special case where the base is $e$).

1. Forgetting the Constant in Growth Models: When solving $e^{kt}=C$, students often write $kt=C$ instead of $kt=\ln C$. You must take the logarithm of the entire other side.

Summary

• Logarithms are the inverse of exponentials. The statement $a^x=y$ is equivalent to $x={\log }_{a}y$.
• The three core laws—${\log }_a(xy)={\log }_{a}x+{\log }_{a}y$, ${\log }_a(x/y)={\log }_{a}x-{\log }_{a}y$, and ${\log }_a(x^k)=k{\log }_{a}x$—are essential for manipulating and solving equations.
• The natural functions $e^x$ and $\ln x$ are a critical pair with unique calculus properties and are central to modeling.
• Exponential growth/decay models ($N=N_0{e}^{kt}$) describe processes with constant relative rates and have direct applications like calculating half-life.
• Logarithmic transformations can linearise exponential data, allowing you to verify a model and find its parameters using linear regression techniques.
• The change of base formula, ${\log }_{a}b=\frac{{\log }_{c}b}{{\log }_{c}a}$, is indispensable for evaluating logs with unfamiliar bases and solving complex exponential equations.