Logarithmic and Exponential Functions

Unlocking Exponents with Logs! 🔑

Introduction

1. 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! ✨

\log_a(xy) = \log_a(x) + \log_a(y)

e

Worked example

Let's Break This Problem Down 🚀

x > 4

1
$First things first, we can't solve this with two separate log terms. We need to combine them into a single logarithm. Let's use the Power Rule on the first term to bring the '2' up as an exponent.$
$2\log_2(x-1) ext{ becomes } \log_2((x-1)^2)$
2
$\log_2((x-1)^2) - \log_2(x+2) = 3$
$\log_2\left(\frac{(x-1)^2}{x+2}\right) = 3$
3
$\log_a(b) = c$
$\frac{(x-1)^2}{x+2} = 2^3 = 8$
4
$We've successfully eliminated the logs! Now it's just an algebra problem. Let's expand the brackets and rearrange everything to form a quadratic equation that we can solve.$
$(x-1)^2 = 8(x+2) \implies x^2 - 2x + 1 = 8x + 16 \implies x^2 - 10x - 15 = 0$
5
$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
$x = \frac{10 \pm \sqrt{(-10)^2 - 4(1)(-15)}}{2(1)} = \frac{10 \pm \sqrt{100 + 60}}{2} = \frac{10 \pm \sqrt{160}}{2}$
6
$x > 4$
$x = \frac{10 \pm \sqrt{16 \times 10}}{2} = \frac{10 \pm 4\sqrt{10}}{2} = 5 \pm 2\sqrt{10} \\ \text{So, } x \approx 11.32 \text{ or } x \approx -1.32 \\ \text{Since } x > 4, \text{ we reject } x = 5 - 2\sqrt{10}. \\ \text{Final Answer: } x = 5 + 2\sqrt{10}$

Answer

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}

a^x = b

p

Worked example

Let's Crack This Code 🕵️‍♀️

A = 800 \times (1.05)^t

y = kx^n

y = k(a^x)

Worked example

Cracking the Code: From Data to Equation 🕵️‍♀️

x

1
$y = kx^n$
$y = kx^n \\ \ln(y) = \ln(kx^n) \\ \ln(y) = \ln(k) + \ln(x^n) \\ \ln(y) = n\ln(x) + \ln(k)$
2
$Y = mX + c$
$m = \frac{\text{change in Y}}{\text{change in X}} = \frac{10.5 - 4.5}{5.0 - 2.0} = \frac{6.0}{3.0} = 2$
3
$m = 2$
$n = m = 2$
4
$Y - Y_1 = m(X - X_1)$
$\ln(y) - 4.5 = 2(\ln(x) - 2.0) \\ \ln(y) = 2\ln(x) - 4.0 + 4.5 \\ \ln(y) = 2\ln(x) + 0.5$
5
$c$
$\ln(k) = 0.5 \\ k = e^{0.5} \\ k \approx 1.65$
6
$We've cracked it! We found both constants. Now we just write out the original equation with our values for k and n . Mission complete.$
$y = 1.65x^2$

Answer

y = 1.65x^2

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