Logarithmic and Exponential Functions | A Level Mathematics (9709) A2 Revision Notes

Introduction

Hey there! Welcome back to the world of functions. Remember how we worked with exponential curves in AS? Well, get ready to take it to the next level. In this chapter, we're going to become absolute masters of logarithms, the secret weapon for taming exponential equations. First, we'll dive deep into the laws of logarithms – the rules that let us manipulate and simplify complex expressions. Then, we'll use these new skills to solve some seriously tricky exponential equations where the unknown is stuck up in the power. Finally, we'll learn a super clever technique called transforming to linear form, which is like a mathematical magic trick for turning curvy graphs into straight lines to make them easier to analyze. Let's get started! ✨

Properties and Laws of Logarithms3.2

Alright, let's talk about logarithms. Think of them as the 'undo' button for exponential functions. If your bank account balance grows exponentially (we wish!), a logarithm could tell you how long it would take to reach a million dollars. The rules that govern them are basically cheat codes that let you simplify complicated-looking expressions. There are three main laws you need to master. First up, the Product Rule: $\log_a(xy) = \log_a(x) + \log_a(y)$. When two numbers are multiplied inside a log, you can split them into two logs being added. It’s like merging two of your Spotify playlists into one big one. Next, the Quotient Rule: $\log_a(\frac{x}{y}) = \log_a(x) - \log_a(y)$. When you're dividing inside a log, you can split it into two logs being subtracted. Think of it like calculating the net change in your Insta followers after a controversial post. Finally, the most clutch rule of all, the Power Rule: $\log_a(x^n) = n \log_a(x)$. This lets you take an exponent from inside the log and drop it down as a multiplier in front. This is a total game-changer for solving equations where the unknown is a power. A clean, colourful graphic summarizing the three laws of logarithms. Product Rule: log(ab) = log(a)+log(b). Quotient Rule: log(a/b) = log(a)-log(b). Power Rule: log(a^n) = n*log(a). Each rule has a simple numerical example. And what about 'ln'? That's just the 'natural logarithm', which is a fancy way of saying a log with a specific base called '$e$' (which is roughly 2.718). It's like the VIP version of logs because it shows up everywhere in advanced maths and science. The good news? All the same rules apply. So, $\ln(x^2) = 2\ln(x)$. Don't forget the basics either: $\log_a(a) = 1$ (and so $\ln(e) = 1$) and $\log_a(1) = 0$ (and so $\ln(1) = 0$). Mastering these rules is all about taking something that looks messy and making it simple enough to solve.

Worked Example: Applying Logarithmic Laws to Solve an Equation3.2

Problem Given that $x > 4$, solve the equation $2\log_2(x-1) - \log_2(x+2) = 3$.

Step 1 First things first, we can't solve this with two separate log terms. We need to combine them into a single logarithm. Let's use the Power Rule on the first term to bring the '2' up as an exponent. $$2\log_2(x-1) ext{ becomes } \log_2((x-1)^2)$$

Step 2 Now our equation is $\log_2((x-1)^2) - \log_2(x+2) = 3$. We have one log subtracting another, which is the perfect setup for the Quotient Rule. Let's combine them into a single log where the arguments are divided. $$\log_2\left(\frac{(x-1)^2}{x+2}\right) = 3$$

Step 3 Okay, mission accomplished! We have a single log. Now we can get rid of it by converting the equation into its exponential form. Remember, $\log_a(b) = c$ is the same as $a^c = b$. Here, our base 'a' is 2, 'c' is 3, and 'b' is that whole fraction. $$\frac{(x-1)^2}{x+2} = 2^3 = 8$$

Step 4 We've successfully eliminated the logs! Now it's just an algebra problem. Let's expand the brackets and rearrange everything to form a quadratic equation that we can solve. $$(x-1)^2 = 8(x+2) \implies x^2 - 2x + 1 = 8x + 16 \implies x^2 - 10x - 15 = 0$$

Step 5 This quadratic doesn't factor nicely, so it's time for the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. Let's plug in our values ($a=1, b=-10, c=-15$). $$x = \frac{10 \pm \sqrt{(-10)^2 - 4(1)(-15)}}{2(1)} = \frac{10 \pm \sqrt{100 + 60}}{2} = \frac{10 \pm \sqrt{160}}{2}$$

Step 6 Let's simplify that radical and find our two possible solutions. Remember the original problem stated that $x > 4$. We need to check if our answers satisfy this condition. The log arguments $(x-1)$ and $(x+2)$ must also be positive. $$x = \frac{10 \pm \sqrt{16 \times 10}}{2} = \frac{10 \pm 4\sqrt{10}}{2} = 5 \pm 2\sqrt{10} \\ \text{So, } x \approx 11.32 \text{ or } x \approx -1.32 \\ \text{Since } x > 4, \text{ we reject } x = 5 - 2\sqrt{10}. \\ \text{Final Answer: } x = 5 + 2\sqrt{10}$$

Solving Exponential Equations Using Logarithms3.2

Alright, let's get into it. You've seen exponential equations before, where things grow ridiculously fast, kind of like a TikTok sound going viral overnight. The equations look like $a^x = b$. The challenge? The unknown, $x$, is stuck up in the exponent, like it's a VIP in the cloud. You can't just divide by 'a' to get it down. So, how do we solve for it? We need a special move, and that move is called logarithms. Think of logarithms as the inverse of exponentials, the same way subtraction is the inverse of addition. They're the only tool that can bring that exponent down to our level. The main strategy is simple: take logs of both sides of the equation. As long as you do the same thing to both sides, the equation stays balanced. You can use any log base, but your calculator's best friends are the natural logarithm, $\ln(x)$ (which is log base $e$), and the common log, $\lg(x)$ (log base 10). We usually stick with $\ln(x)$ in A-Levels because the number $e$ is a massive deal in maths. Once you take logs, you unleash the ultimate power-up: the power law of logs. This law says $\log(M^p) = p \log(M)$. A diagram showing the exponent 'p' in log(M^p) moving down in front of the log to become p log(M), with arrows indicating the movement. See what happened? The exponent $p$ is now a regular multiplier. Our equation $a^x = b$ becomes $\ln(a^x) = \ln(b)$, and using the power law, it transforms into $x \ln(a) = \ln(b)$. Suddenly, it's just a simple linear equation you've been solving for years! Just divide by $\ln(a)$ and you've got your answer: $x = \frac{\ln(b)}{\ln(a)}$. The same idea applies to inequalities like $3^x > 100$. You solve it the same way, just be mindful that if you ever divide by a negative log (which happens if the base is between 0 and 1), you'd have to flip the inequality sign. It's your secret weapon for finding out exactly when your investment hits a target or how long a radioactive substance takes to decay.

Worked Example: Solving an Exponential Equation3.2

Problem Your savings account for a new gaming PC grows according to the formula $A = 800 \times (1.05)^t$, where $t$ is the number of years. How many full years will it take for your savings to first exceed $1200? Give your answer as an integer.

Step 1 First, we set up the inequality. We want to find the time $t$ when the amount $A$ is greater than 1200. Then, we'll isolate the exponential part by dividing both sides by 800. $$ 800 \times (1.05)^t > 1200 \\ (1.05)^t > \frac{1200}{800} \\ (1.05)^t > 1.5 $$

Step 2 Now the variable $t$ is stuck in the exponent. This is our cue to use logarithms! We'll take the natural logarithm (ln) of both sides to bring the power down. $$ \ln((1.05)^t) > \ln(1.5) $$

Step 3 Time for the magic trick: apply the power law of logarithms, $\ln(M^p) = p \ln(M)$. This brings the $t$ down from the exponent, turning it into a coefficient. $$ t \ln(1.05) > \ln(1.5) $$

Step 4 We can now solve for $t$ by dividing both sides by $\ln(1.05)$. Since $1.05 > 1$, we know $\ln(1.05)$ is a positive number, so we don't need to flip the inequality sign. Phew! $$ t > \frac{\ln(1.5)}{\ln(1.05)} $$

Step 5 Use your calculator to find the decimal value. Remember to keep a few extra decimal places for accuracy before the final rounding. $$ t > \frac{0.405465...}{0.048790...} \\ t > 8.310... $$

Step 6 The question asks for the number of full years it takes to first exceed $1200. Since $t$ must be greater than 8.31, the first integer value for $t$ that satisfies this is 9. So, it will take 9 full years. $$ t = 9 \text{ years} $$

Transforming Non-Linear Relationships to Linear Form3.2

Alright, let's be real. Straight lines are predictable and easy to work with. You know the gradient, you know the intercept, life is good. It's like knowing exactly how much data your music streaming will use per hour. But a lot of relationships in science, finance, and even gaming aren't that simple. They're curves. Trying to analyze a curve directly is like trying to predict the next viral TikTok dance – chaotic and complicated. So, what's the hack? We give the equation a 'glow-up' and transform it into a linear one. This whole process is called linearisation, and it's a total game-changer. We focus on two main types of relationships. First, the power law form: $y = kx^n$. Think about something like the relationship between a car's speed and its stopping distance. To straighten this out, we take logs of both sides (you can use natural logs, $\ln$, or logs to base 10, $\lg$). Let's use $\ln$: $\ln(y) = \ln(kx^n)$ Using our log laws (remember those?), we can split this up: $\ln(y) = \ln(k) + \ln(x^n)$ And bring the power down: $\ln(y) = n\ln(x) + \ln(k)$ Now, look closely. This looks exactly like the equation of a straight line, $Y = mX + c$. We've just made a substitution! Here, $Y = \ln(y)$, $X = \ln(x)$, the gradient $m = n$, and the y-intercept $c = \ln(k)$. So if you plot a graph of $\ln(y)$ against $\ln(x)$, you'll get a perfect straight line. From that graph, you can find the gradient and intercept, which lets you find the original secret constants, n and k! A side-by-side diagram. The left graph shows a curve for y=kxⁿ on standard x-y axes. The right graph shows a straight line on axes labelled ln(x) and ln(y), with the gradient 'n' and y-intercept 'ln(k)' clearly indicated. The second type is the exponential form: $y = k(a^x)$. This is huge for things like population growth or your savings account balance. The process is similar. Take logs of both sides: $\ln(y) = \ln(k(a^x)) = \ln(k) + \ln(a^x)$ $\ln(y) = x(\ln(a)) + \ln(k)$ Again, compare this to $Y = mX + c$. This time, $Y = \ln(y)$, but notice that $X = x$ (not $\ln(x)$!). The gradient is $m = \ln(a)$ and the y-intercept is $c = \ln(k)$. So, by plotting $\ln(y)$ against x, you can find the constants a and k. It's basically a clever way to 'un-curve' the data to reveal the simple straight-line relationship hidden inside.

Worked Example: Determining Constants from Experimental Data3.2

Problem Two variables, $x$ and $y$, are believed to be related by the equation $y = kx^n$, where $k$ and $n$ are constants. Experimental data is collected. A graph of $\ln(y)$ against $\ln(x)$ is plotted, resulting in a straight line that passes through the points $(2.0, 4.5)$ and $(5.0, 10.5)$. Determine the values of $k$ and $n$, correct to 3 significant figures.

Step 1 First, let's get our game plan. We need to linearise the given equation $y = kx^n$ by taking natural logs. This will help us compare it to the standard straight-line equation, $Y = mX + c$. $$y = kx^n \\ \ln(y) = \ln(kx^n) \\ \ln(y) = \ln(k) + \ln(x^n) \\ \ln(y) = n\ln(x) + \ln(k)$$

Step 2 By comparing our equation with $Y = mX + c$, we can see that if we plot $Y=\ln(y)$ against $X=\ln(x)$, the gradient m will be equal to n, and the Y-intercept c will be equal to $\ln(k)$. Our first mission is to find the gradient of the line using the two points we're given. $$m = \frac{\text{change in Y}}{\text{change in X}} = \frac{10.5 - 4.5}{5.0 - 2.0} = \frac{6.0}{3.0} = 2$$

Step 3 Nice! We've found the gradient, $m = 2$. Since we know that $m = n$, we've already found our first constant. $$n = m = 2$$

Step 4 Now to find the Y-intercept, c. We can use the point-slope form of a line, $Y - Y_1 = m(X - X_1)$, and sub in one of our points, let's use $(2.0, 4.5)$. Remember, $Y$ is $\ln(y)$ and $X$ is $\ln(x)$. $$\ln(y) - 4.5 = 2(\ln(x) - 2.0) \\ \ln(y) = 2\ln(x) - 4.0 + 4.5 \\ \ln(y) = 2\ln(x) + 0.5$$

Step 5 From that equation, we can see the Y-intercept, $c$, is $0.5$. We also know that $c = \ln(k)$. So, to find k, we need to 'undo' the natural log by using the exponential function, $e^x$. $$\ln(k) = 0.5 \\ k = e^{0.5} \\ k \approx 1.65$$

Step 6 We've cracked it! We found both constants. Now we just write out the original equation with our values for k and n. Mission complete. $$y = 1.65x^2$$

Key Points to Remember

The Laws of Logarithms are your best friends: remember $log_a(x) + log_a(y) = log_a(xy)$, $log_a(x) - log_a(y) = log_a(x/y)$, and $n log_a(x) = log_a(x^n)$. They are your toolkit for everything else!
To solve an exponential equation like $a^{kx} = b$, the main strategy is to take logs of both sides (usually the natural log, $ln$) to bring the power down and isolate the variable.
Transforming to linear form means taking logs of an equation like $y = ax^n$ to get $ln(y) = n ln(x) + ln(a)$, which matches the straight-line equation $Y = mX + c$.

← Chapter 3: Advanced Algebraic Methods P3 Trigonometry: Identities and Harmonic Form →