Logarithms and Exponential Equations for ESAT

Understanding the Nature of Logarithms

Logarithms are fundamentally the inverse of indices. While indices involve raising a base to a power to find a result, logarithms tell you what power a base must be raised to in order to reach a specific value. Historically, logarithms were developed to simplify complex calculations, transforming multiplicative processes into additive ones.

To understand this practically, consider base 10. The expression $\log_{10} 100$ asks: 'To what power must 10 be raised to get 100?' Since $10^2 = 100$, we know $\log_{10} 100 = 2$. Similarly:

1. $\log_{10} 10 = 1$ because $10^1 = 10$.
2. $\log_{10} 1000 = 3$ because $10^3 = 1000$.
3. $\log_{10} 27 \approx 1.431$ because $10^{1.431...} = 27$.

This logic applies to any positive base. For example, in base 2:

1. $\log_2 32 = 5$ because $2^5 = 32$.
2. $\log_2 \frac{1}{2} = -1$ because $2^{-1} = \frac{1}{2}$.

In general, the relationship is defined as: $\log_a c = b \Leftrightarrow a^b = c$. Note the following constraints: the base $a$ must be positive and not equal to 1, and the argument $c$ must be positive. However, the logarithm itself (the exponent $b$) can be any real number, including negative values or zero.

Graphical Representation of Logarithms

Because logarithms are the inverse of exponential functions, their graphs are reflections of each other. Consider the function $y = 2^x$. This function takes a value $x$ and maps it to $2^x$ on the $y$-axis.

If we start on the $y$-axis with a value (for instance, 8) and trace back to the corresponding $x$-value (3), we are performing a logarithmic operation: $\log_2 8 = 3$. The graph of the logarithm $y = \log_2 x$ is simply the graph of $y = 2^x$ with the $x$ and $y$ axes swapped.

Observation of the log graph reveals that it is only defined for $x > 0$. It also crosses the $x$-axis at 1, because $2^0 = 1$, which implies $\log_2 1 = 0$ for any valid base.

The Laws of Logarithms

You must be able to use the following algebraic laws, which are derived from the laws of indices. The identity $a^{\log_a x} = x$ is central to these proofs.

1. Addition Law: $\log_a x + \log_a y = \log_a(xy)$. This corresponds to the index law $a^p a^q = a^{p+q}$.
2. Subtraction Law: $\log_a x - \log_a y = \log_a(\frac{x}{y})$. This corresponds to $a^p / a^q = a^{p-q}$.
3. Power Law: $k \log_a x = \log_a(x^k)$. This corresponds to $(a^p)^k = a^{pk}$.

Special cases derived from these include $\log_a (\frac{1}{x}) = -\log_a x$ and the identity $\log_a a = 1$.

Change of Base Formula

Although the change of base formula is not explicitly tested in the ESAT, it is a powerful tool. It allows you to convert a logarithm from one base to another: $\log_a b = \frac{\log_c b}{\log_c a}$. This leads to the useful reciprocal identity $\log_a b = \frac{1}{\log_b a}$.

Solving Exponential Equations

Equations of the form $a^x = b$ are solved by taking logs of both sides. It is often best to keep answers in exact form using logarithms rather than rounding decimals.

Worked Example: Solve $5^{2x} = 27$

Approach 1: Using base 5 Take $\log_5$ of both sides: $\log_5 5^{2x} = \log_5 27$. Using the power law and the fact that $27 = 3^3$, we get $2x = 3 \log_5 3$. Thus, $x = \frac{3}{2} \log_5 3$.

Approach 2: Using base 3 Take $\log_3$ of both sides: $\log_3 5^{2x} = \log_3 27$. Since $\log_3 27 = 3$, we have $2x \log_3 5 = 3$. This gives $x = \frac{3}{2 \log_3 5}$. Both approaches yield identical values.

More complex equations can often be reduced to this form via substitution. For example, $25^x - 3 \times 5^x + 2 = 0$ can be rewritten as $(5^x)^2 - 3(5^x) + 2 = 0$. Letting $u = 5^x$, we solve the quadratic $u^2 - 3u + 2 = 0$ to find $u = 1$ or $u = 2$, and then solve $5^x = 1$ and $5^x = 2$ for $x$.