The Use of Logarithms with GPs Revision Notes for HSC SSCE Mathematics Advanced

The Use of Logarithms with GPs

Introduction

Finding the number of terms in a geometric progression (GP) requires solving equations where the variable $n$ appears in the exponent. There are two main approaches to solving these exponential equations:

Trial-and-error: This method involves testing different values of $n$ on a calculator until you find the answer. While easier to understand initially, it becomes inefficient when working with large numbers.

Logarithms: This provides a more sophisticated and efficient approach, but requires understanding the relationship between logarithms and indices.

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The choice between trial-and-error and logarithms often depends on the context. For small values, trial-and-error might be quick enough, but for larger problems, logarithms become essential.

Solving exponential inequations using trial-and-error

Exponential inequations involve equations with $n$ in the index position. The trial-and-error method uses a calculator to test successive values until the correct answer is found.

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Worked Example: Trial-and-error method

Question: Use trial-and-error on your calculator to find the smallest integer $n$ such that:

a) $3^n > 400000$

b) $0.95^n < 0.01$

Solution:

Part a:

Using the power function on your calculator:

3^{11} = 177147 3^{12} = 531441

So the smallest such integer is 12.

Part b:

Using the power function on your calculator:

0.95^{89} = 0.010408\ldots 0.95^{90} = 0.009888\ldots

So the smallest such integer is 90.

chatImportant

Critical Observation: Notice how the powers of $3$ get bigger because the base 3 is greater than 1. The powers of $0.95$ , however, get smaller because the base 0.95 is less than 1. This behavior is crucial for solving exponential inequations correctly.

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In practice, quite a few more trial calculations are usually needed to trap the given number between two integer powers. This is why logarithms become more efficient for complex problems.

Application: Inflation and GPs

The following example demonstrates how geometric progressions model real-world scenarios like price inflation over time.

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Worked Example: Widget Company Sales

Question: The General Widget Company has sold $2000$ widgets per year since its foundation in $2011$ when the company charged $300 per widget. Each year, the company lifts its prices by $5\%$ because of cost increases.

a) Find the value of the sales in the $n$ th year after $2010$ .

b) Using trial-and-error, find the first year in which sales exceeded $900000.

c) Find the total sales from the foundation of the company to the end of the $n$ th year.

d) Using trial-and-error, find the year during which the total sales of the company since its foundation will first exceed $20000000.

Solution:

The value of the annual sales in $2011$ were $300 $\times$ $2000$ = $600000.

Hence the annual sales form a GP with $a = 600000$ and $r = 1.05$ .

Part a:

The sales in the $n$ th year after $2010$ constitute the $n$ th term $T_n$ of the series.

$T_n = ar^{n-1}$

$T_n = 600000 \times 1.05^{n-1}$

Part b:

Sales in $2019$ = $T_9$

$= 600000 \times 1.05^8$

$\approx \text{\$886473}$

Sales in $2020$ = $T_{10}$

$= 600000 \times 1.05^9$

$\approx \text{\$930797}$

Hence the sales first exceeded $900000 in the year 2020.

Part c:

The total sales since the foundation of the company constitute the sum $S_n$ of the first $n$ terms of the series.

$S_n = \frac{a(r^n - 1)}{r - 1} \quad \text{(using this formula because } r > 1\text{)}$

$S_n = \frac{600000 \times (1.05^n - 1)}{0.05}$

$S_n = 12000000 \times (1.05^n - 1)$

Part d:

Total sales to $2030$ = $S_{20}$

$= 12000000 \times (1.05^{20} - 1)$

$\approx \text{\$19839572}$

Total sales to $2031$ = $S_{21}$

$= 12000000 \times (1.05^{21} - 1)$

$\approx \text{\$21431551}$

Hence cumulative sales will first exceed $20000000 during 2031.

Using logarithms to solve exponential inequations

To solve an exponential inequation using logarithms, follow these two key steps:

Step 1: Convert the exponential equation to a logarithmic equation.

Step 2: Since calculators only have logarithms base $10$ and $e$ , use the change-of-base formula:

$\log_b x = \frac{\log_{10} x}{\log_{10} b} \quad \text{OR} \quad \log_b x = \frac{\log_e x}{\log_e b}$

infoNote

Memory aid: "The log of the number over log of the base."

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Worked Example: Using Logarithms

Question: Use logarithms to find the smallest integer $n$ such that:

a) $3^n > 400000$

b) $1.04^n > 2$

Solution:

Part a:

Put $3^n = 400000$ (beginning with the corresponding equation)

Then $n = \log_3 400000$ (converting to a logarithmic equation)

Using the change-of-base formula:

$n = \frac{\log_{10} 400000}{\log_{10} 3} \quad \text{OR} \quad \frac{\log_e 400000}{\log_e 3}$

$n = 11.741...$

Thus the smallest such integer is 12, because $3^{11} < 400000$ and $3^{12} > 400000$ .

Part b:

Put $1.04^n = 2$

Then $n = \log_{1.04} 2$

$n = \frac{\log_{10} 2}{\log_{10} 1.04} \quad \text{OR} \quad \frac{\log_e 2}{\log_e 1.04}$

$n = 17.672...$

Thus the smallest such integer is 18, because $1.04^{17} < 2$ and $1.04^{18} > 2$ .

An alternative approach: Taking logarithms of both sides

There is an equally effective alternative method where you take logarithms base 10 or base e of both sides, then use the logarithmic law:

$\log_{10} a^n = n \log_{10} a \quad \text{OR} \quad \log_e a^n = n \log_e a$

infoNote

This law states: "The log of the power is the multiple of the log."

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Worked Example: Alternative Method

Question: Use logarithms to find the smallest integer $n$ such that:

a) $3^n > 400000$

b) $1.04^n > 2$

Solution:

Part a:

Put $3^n = 400000$ (beginning with the corresponding equation)

Then $\log_{10} 3^n = \log_{10} 400000$ (taking logarithms base 10 of both sides)

$n \log_{10} 3 = \log_{10} 400000 \quad \text{(the log of a power is the multiple of the log)}$

$n = \frac{\log_{10} 400000}{\log_{10} 3} \quad \text{(dividing both sides by } \log_{10} 3\text{)}$

$n = 11.741...$

Thus the smallest such integer is 12, because $3^{11} < 400000$ and $3^{12} > 400000$ .

Part b:

Put $1.04^n = 2$

Then $\log_{10} 1.04^n = \log_{10} 2$

$n \log_{10} 1.04 = \log_{10} 2$

$n = \frac{\log_{10} 2}{\log_{10} 1.04}$

$n = 17.672...$

Thus the smallest such integer is 18, because $1.04^{17} < 2$ and $1.04^{18} > 2$ .

infoNote

This alternative approach takes one more line of working and involves, in effect, a proof of the change-of-base formula. You can use either method - choose the one you find clearer. You can also use logarithms base $e$ instead of base $10$ .

Applying logarithms to problems

Logarithms are essential for solving real-world GP problems where you need to find which term reaches a certain value. The following example demonstrates this application in a business context.

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Worked Example: Business Profits

Question: The profits of the Extreme Sports Adventure Company have been increasing by $15\%$ every year since its formation, when its profit was $60000 in the first year.

a) Find a formula for its profit in the $n$ th year.

b) During which year did its profit first exceed $1200000?

c) Find a formula for its total profit during the first $n$ years.

d) During which year did its total profit since foundation first exceed $4000000?

Solution:

The successive profits form a GP with $a = 60000$ and $r = 1.15$ .

Part a:

Profit in the $n$ th year = $T_n$

$T_n = ar^{n-1}$

$T_n = 60000 \times 1.15^{n-1}$

Part b:

Put $T_n = 1200000$ (the corresponding equation)

Then $60000 \times 1.15^{n-1} = 1200000$

Divide by $60000$ :

$1.15^{n-1} = 20$

$n - 1 = \log_{1.15} 20$

$n - 1 = \frac{\log_{10} 20}{\log_{10} 1.15} \quad \text{or} \quad \frac{\log_e 20}{\log_e 1.15}$

$n - 1 \approx 21.43$

Add $1$ :

$n \approx 22.43$

So the profit first exceeds $1200000 during the 23rd year.

Part c:

Total profit in the first $n$ years = $S_n$

$S_n = \frac{a(r^n - 1)}{r - 1} \quad \text{(using this form because } r > 1\text{)}$

$S_n = \frac{60000 \times (1.15^n - 1)}{0.15}$

$S_n = 400000 \times (1.15^n - 1)$

Part d:

To find the year in which total profit first exceeded $4000000:

Put $S_n = 4000000$ (the corresponding equation)

Then $400000 \times (1.15^n - 1) = 4000000$

Divide by $400000$ :

$1.15^n - 1 = 10$

$1.15^n = 11$

$n = \log_{1.15} 11$

$n = \frac{\log_{10} 11}{\log_{10} 1.15} \quad \text{OR} \quad \frac{\log_e 11}{\log_e 1.15}$

$n \approx 17.16$

So the total profit since foundation first exceeds $4000000 during the 18th year.

Using logarithms when the base is less than 1

Understanding how the base affects the sequence is crucial:

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When the base is greater than 1: The successive powers form an increasing sequence.

For example, the powers of $2$ are:

$2^1 = 2, \quad 2^2 = 4, \quad 2^3 = 8, \quad 2^4 = 16, ...$

When the base is less than 1: The successive powers form a decreasing sequence.

For example, the powers of $\frac{1}{2}$ are:

$\left(\frac{1}{2}\right)^1 = \frac{1}{2}, \quad \left(\frac{1}{2}\right)^2 = \frac{1}{4}, \quad \left(\frac{1}{2}\right)^3 = \frac{1}{8}, \quad \left(\frac{1}{2}\right)^4 = \frac{1}{16}, ...$

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Therefore, when the base is less than one, questions will ask for the smallest value of the index that makes the power less than some small number (not greater than).

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Worked Example: Base Less Than 1

Question: Use logarithms to find the smallest value of $n$ such that:

a) $\left(\frac{1}{3}\right)^n < 0.000001$

b) $0.95^n < 0.01$

Solution:

Part a:

Put $\left(\frac{1}{3}\right)^n = 0.000001$ (the corresponding equation)

Then $n = \log_{\frac{1}{3}} 0.000001$ (converting to a logarithmic equation)

$n = \frac{\log_{10} 0.000001}{\log_{10} \frac{1}{3}} \quad \text{(using the change-of-base formula)}$

$n = 12.575...$

Thus the smallest such integer is 13, because $\left(\frac{1}{3}\right)^{12} > 0.000001$ and $\left(\frac{1}{3}\right)^{13} < 0.000001$ .

Part b:

Put $0.95^n = 0.01$

Then $n = \log_{0.95} 0.01$

$n = \frac{\log_{10} 0.01}{\log_{10} 0.95} \quad \text{OR} \quad \frac{\log_e 0.01}{\log_e 0.95} \quad \text{(change-of-base formula)}$

$n = 89.781...$

Thus the smallest such integer is 90, because $0.95^{89} > 0.01$ and $0.95^{90} < 0.01$ .

Application: Declining sales

The following example demonstrates how to handle geometric progressions with a decreasing ratio in real-world scenarios.

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Worked Example: Declining Sales

Question: Consider the slowly failing Gumnut Softdrinks Factory in Wadelbri, where sales are declining by $6\%$ every year, with $50000$ bottles sold in $2016$ .

During which year will sales first fall below $20000$ ?

Solution:

The sales form a GP with $a = 50000$ and $r = 0.94$ .

Put $T_n = 20000$ (this is the corresponding equation)

Then $ar^{n-1} = 20000$

$50000 \times 0.94^{n-1} = 20000$

Divide by $50000$ :

$0.94^{n-1} = 0.4$

$n - 1 = \log_{0.94} 0.4 \quad \text{(converting to a logarithmic equation)}$

$n - 1 = \frac{\log_{10} 0.4}{\log_{10} 0.94} \quad \text{OR} \quad \frac{\log_e 0.4}{\log_e 0.94} \quad \text{(change-of-base)}$

$n - 1 \approx 14.81$

Add $1$ :

$n \approx 15.81$

Hence sales will first fall below $20000$ when $n = 16$ , that is, in 2031.

Key method: Solving exponential inequations

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Key Approaches to Solving Exponential Inequations

To solve an exponential inequation such as $3^n > 400000$ or $0.95^n < 0.01$ :

Approach 1: Trial-and-error

Use the calculator to test successive values
Stop when you trap the number between two consecutive integer powers

Approach 2: Using logarithms

Write down the corresponding equation: $3^n = 400000$ or $0.95^n = 0.01$
Solve for $n$ , giving $n = \log_3 400000$ or $n = \log_{0.95} 0.01$
Convert to logarithms base $e$ or base $10$ using the change-of-base formula
Approximate the value
Write down the solution of the corresponding inequation

Approach 3: Taking logarithms of both sides

Take logarithms base $e$ or base $10$ of both sides
Use the log law: $\log a^n = n \log a$
Solve for $n$

chatImportant

Critical Point: Remember that powers of bases greater than 1 get bigger as the index increases, while powers of bases less than 1 get smaller as the index increases.

Remember!

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Key Points to Remember:

When finding the number of terms in a GP, you must solve an exponential equation where $n$ is in the index
The change-of-base formula is essential: $\log_b x = \frac{\log_{10} x}{\log_{10} b}$ (remember: "log of the number over log of the base")
You can either use the change-of-base formula directly or take logarithms of both sides - both methods work equally well
Always start by writing the corresponding equation before solving an inequation
For bases greater than 1: powers increase, so you need $n$ such that $a^n > k$
For bases less than 1: powers decrease, so you need $n$ such that $a^n < k$
When solving practical problems, always round up to the next whole number for "during which year" questions

The Use of Logarithms with GPs (HSC SSCE Mathematics Advanced): Revision Notes