Logarithms in A Level Maths | 5 Best Examples
Logarithms in A Level Maths – Introduction
In A Level Maths, logarithms are essential tools that help solve equations involving exponential functions and simplify complex mathematical calculations. This blog post aims to provide a clear understanding of what logarithms are and how they are applied in the context of A Level Maths.
What are Logarithms?
Logarithms are mathematical functions that represent the inverse operation of exponentiation. They are used to solve equations where the variable appears as an exponent. Logarithms essentially ask the question, To what power must a given base be raised to obtain a certain value?
Basic Logarithmic Properties
Logarithmic Notation: Logarithms are denoted using the base and the argument. The general form is written as log(base)(argument). For example, log₂(8) represents the logarithm of 8 to the base 2.
Change of Base Formula: When the base of a logarithm is not specified, it is usually assumed to be 10. However, logarithms can be calculated with any base. The change of base formula allows us to convert logarithms from one base to another. The formula is as follows: log(base₁)(x) = log(base₂)(x) / log(base₂)(base₁).
Product Rule: When multiplying two numbers with the same base, the logarithm of the product is equal to the sum of the logarithms of the individual numbers. Mathematically, it can be expressed as log(base)(xy) = log(base)(x) + log(base)(y).
Quotient Rule: When dividing two numbers with the same base, the logarithm of the quotient is equal to the difference of the logarithms of the individual numbers. Mathematically, it can be expressed as log(base)(x/y) = log(base)(x) – log(base)(y).
Applications of Logarithms
Solving Exponential Equations: Logarithms are particularly useful when solving equations where the variable appears as an exponent. By applying logarithmic operations, we can isolate the variable and determine its value. For example, if we have the equation 2^x = 16, we can rewrite it as x = log(base 2)(16).
Compound Interest and Half-Life: Logarithms are also applied in various real-world scenarios, such as compound interest and radioactive decay. In compound interest calculations, logarithms help determine the time required for an investment to reach a certain value. Similarly, in half-life calculations, logarithms are used to find the time it takes for a substance to decay to half its initial amount.
Data Compression: Logarithms are utilised in data compression algorithms, where they help reduce the size of data files without significant loss of information. By applying logarithmic functions, unnecessary or redundant information can be compressed, resulting in smaller file sizes.
Logarithms – Example 1
Solution
a) 3^2=9, \text { so } \log _3 9=2
b) 2^7=128, \text { so } \log _2 128=7
c) 64 \frac{1}{2}=8,50 \log _{64} 8=\frac{1}{2}
The Laws of Logarithms
Logarithms – Example 2
a) \begin{aligned} & \log _3(6 \times 7) \\ & =\log _3 42 \end{aligned}
b) \begin{aligned} & \log _2(15 \div 3) \\ & =\log _2 5 \end{aligned}
c) \begin{aligned} & 2 \log _5 3=\log _5\left(3^2\right)=\log _5 9 \\ & 3 \log _5 2=\log _5\left(2^3\right)=\log _5 8 \\ & \log _5 9+\log _5 8=\log _5 72 \end{aligned}
d) 4 \log _{10}\left(\frac{1}{2}\right)=\log _{10}\left(\frac{1}{2}\right)^4=\log _{10}\left(\frac{1}{16}\right)
\log _{10} 3-\log _{10}\left(\frac{1}{16}\right)=\log _{10}\left(3 \div \frac{1}{16}\right)=\log _{10} 48Logarithms – Example 3
Solution
a) \begin{aligned} & \log _a\left(x^2 y z^3\right) \\ & =\log _a\left(x^2\right)+\log _a y+\log _a\left(z^3\right) \\ & =2 \log _a x+\log _a y+3 \log _a z \end{aligned}
b) \begin{aligned} & \log _a\left(\frac{x}{y^3}\right) \\ & =\log _a x-\log _a\left(y^3\right) \\ & =\log _a x-3 \log _a y \end{aligned}
c) \begin{aligned} & \log _a\left(\frac{x \sqrt{y}}{z}\right) \\ & =\log _a(x \sqrt{y})-\log _a z \\ & =\log _a x+\log _a \sqrt{y}-\log _a z \\ & =\log _a x+\frac{1}{2} \log _a y-\log _a z \end{aligned}
d) \begin{aligned} & \log _a\left(\frac{x}{a^4}\right) \\ & =\log _a x-\log _a\left(a^4\right) \\ & =\log _a x-4 \log _a a \\ & =\log _a x-4 \end{aligned}
Logarithms – Example 4
Solution
\begin{aligned} & \log _{10} 4+2 \log _{10} x=2 \\ & \log _{10} 4+\log _{10} x^2=2 . \\ & \log _{10} 4 x^2=2 . \\ & 4 x^2=10^2 \\ & 4 x^2=100 \\ & x^2=25 \\ & x=5 \end{aligned}Logarithms – Example 5
Conclusion
Logarithms play a crucial role in A Level Maths, providing a powerful tool for solving exponential equations and simplifying complex calculations. Understanding the basic properties of logarithms and their applications can greatly enhance problem-solving abilities in various mathematical contexts.