Geometric Sequences with Logarithms

Expert reviewed 21 July 2024 6 minute read


HSC Maths Advanced Syllabus

  • use geometric sequences to model and analyse practical problems involving exponential growth and decay
    • calculate the effective annual rate of interest and use results to compare investment returns and cost of loans when interest is paid or charged daily, monthly, quarterly or six-monthly

Note:

Video coming soon!

Logarithms and Geometric Sequences

Logarithms are frequently used in finance, particularly in conjunction with geometric sequences, to solve problems involving compound interest, growth rates, and the value of money. Here is an in-depth explanation of how solve logarithmic and exponential inequalities:

The first method is to simply use trial and error to solve the inequalities. This generally works when small values are used. For example, for the inequality 2n<702^n<70, we can simply substitute small values of nn, to see what values result in values less than 7070.

Thus, we can substitute in n=6n=6, as a rough estimate:

26<7064<702^6 <70\\64<70

Therefore, we can see that n=6n=6 fits the inequality as 6464 is less than 7070 . However, we also know that it is the largest integer that fits the inequality. It should be noted that as numbers grow in value, the accuracy of this method is diminished.

The second method is to turn the inequality into an equation, solve for the unknown variable, and then depending on the original inequality sign, determine the closest integer which fits the inequality. Let’s complete these steps using the example, 2n<5000002^n<500000.

Turning the inequality into an equation we get:

2n=5000002^n=500000

Now we solve the equation to find nn:

n=log2500000=log500000log2=18.93n=log_2500000\\=\frac{log500000}{log2}\\=18.93

Looking at the original inequality, we know that 2n<5000002^n< 500000, and as such the largest possible integer nn can equal to fit this inequality, is 1818.

Practice Question 1

The profits of the "High Altitude Exploration Company" have been increasing by 20% every year since its formation, when its profit was $80,000 in the first year. During which year did its profit first exceed $500 000?

The profits follow a geometric sequence where the first term is a=80000a=80000 and the common ration r=1.2r=1.2. Now using, the formula to find the nnth term of a geometric sequence, we can determine nn.

Tn=arnT_n=ar^{n}\\

In this case, the formula has change slightly as we are determining a specific year, thus what is usually (n1)(n-1) in the power, is now just nn.

arn>50000080000(1.2)n>500001.2n>6.25n>log1.26.25n>ln6.25ln1.2n>10.051...ar^n>500000\\80000(1.2)^n>50000\\1.2^n>6.25 \\n>log_{1.2}6.25\\n>\frac{ln6.25}{ln1.2}\\n>10.051...

\therefore Since nn must be a whole number, we round up to the nearest whole year, thus n=11n=11

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