Solving Exponential and Logarithmic Equations for the ESAT

Foundations of Logarithms

Logarithms are closely related to indices: they are the inverse of indices. Instead of raising a base to a power to find a result, a logarithm tells you what power a base must be raised to in order to reach a specific number. Before the invention of calculators, logarithms were essential for simplifying complex calculations with large numbers.

Consider base 10 examples:

1. $\log_{10} 10 = 1$ because $10^1 = 10$.
2. $\log_{10} 100 = 2$ because $10^2 = 100$.
3. $\log_{10} 1000 = 3$ because $10^3 = 1000$.
4. $\log_{10} 27 = 1.431363764\dots$ because $10^{1.431363764\dots} = 27$.

We can use other bases, such as base 2:

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

General definition: $\log_a c = b$ is equivalent to $a^b = c$. There are three critical conditions for this relationship in standard mathematics:

• The base must be positive: $a > 0$ and $a \neq 1$.
• You can only take the log of a positive number: $c > 0$.
• The result of a log can be any number: $b$ can be positive, negative, or zero.

Logarithms and Graphs

The relationship between exponentials and logarithms is clearly visible when graphed. Using base 2 as an example, we can see the exponential growth of $y = 2^x$.

The function $2^x$ maps a value from the $x$ axis to the $y$ axis. If we start on the $y$ axis and trace back to find the corresponding $x$ value, we are performing the log operation: $y \to \log_2 y$. By swapping the $x$ and $y$ axes, we produce the graph of $y = \log_2 x$.

Note that the graph is only defined for $x > 0$. It crosses the $x$ axis at 1 because $2^0 = 1$, which means $\log_2 1 = 0$.

Laws of Logarithms

You must be able to use the following laws to manipulate equations:

1. The Product Law: $\log_a x + \log_a y = \log_a(xy)$. This is derived from $a^{\log_a x + \log_a y} = a^{\log_a x} a^{\log_a y} = xy = a^{\log_a(xy)}$. Note that $a^{\log_a x} = x$ by definition.
2. The Quotient Law: $\log_a x - \log_a y = \log_a \left(\frac{x}{y}\right)$. This is derived from $a^{\log_a x - \log_a y} = \frac{a^{\log_a x}}{a^{\log_a y}} = \frac{x}{y}$.
3. The Power Law: $k \log_a x = \log_a(x^k)$. This is derived from $a^{k \log_a x} = (a^{\log_a x})^k = x^k$.
4. Reciprocal Case: $\log_a \frac{1}{x} = -\log_a x$.
5. Identity Case: $\log_a a = 1$ because $a^1 = a$.

Solving $a^x = b$

To solve equations where the variable is in the index, we take logs of both sides. Most log values are irrational, so we often leave answers in exact form rather than rounding to decimals.

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

Approach 1 (Base 5): Take $\log_5$ of both sides: $\log_5 5^{2x} = \log_5 27$. Since $27 = 3^3$, we have $2x = \log_5 3^3$. Using the power law: $2x = 3 \log_5 3$. Dividing by 2: $x = \frac{3}{2} \log_5 3$.

Approach 2 (Base 3): Take $\log_3$ of both sides: $\log_3 5^{2x} = \log_3 27$. Using the power law: $2x \log_3 5 = 3$ (since $\log_3 27 = 3$). Dividing: $x = \frac{3}{2 \log_3 5}$.

Equations Reducible to Linear or Quadratic Form

Some equations require algebraic manipulation before they can be solved. A common type is the hidden quadratic, such as $25^x - 3 \times 5^x + 2 = 0$.

Recognise that $25^x = (5^2)^x = (5^x)^2$. If we let $u = 5^x$, the equation becomes $u^2 - 3u + 2 = 0$. Factoring gives $(u-2)(u-1) = 0$, so $u = 2$ or $u = 1$. Substituting back: $5^x = 2$ (so $x = \log_5 2$) or $5^x = 1$ (so $x = 0$).

The Change of Base Formula

While questions requiring the change of base formula specifically will not be set, it is useful for your mathematical toolkit. It allows you to convert from base $a$ to base $c$: $\log_a b = \frac{\log_c b}{\log_c a}$

Derivation: Let $p = \log_a b$, so $a^p = b$. Take $\log_c$ of both sides: $\log_c a^p = \log_c b$. This gives $p \log_c a = \log_c b$, so $p = \frac{\log_c b}{\log_c a}$. A special case occurs when $c = b$, leading to: $\log_a b = \frac{1}{\log_b a}$.