Solving Exponential and Logarithmic Equations

Equations involving exponents, such as $a^x = b$, require the use of logarithms to isolate the unknown variable $x$. In the TMUA, you are often expected to provide exact solutions. This means expressing the final answer using logarithmic terms or surds rather than calculating a decimal approximation. Many logarithmic values are irrational, and using exact forms preserves mathematical precision.

Solving Basic Exponential Equations

The most direct method to solve $a^x = b$ is to take the logarithm of both sides. While any base can be used, it is often most efficient to use the base of the exponential term or a base common to both sides of the equation. This allows you to apply the laws of logarithms, specifically the power rule: $\log_a (x^k) = k \log_a x$.

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

There are multiple ways to solve this equation exactly depending on the base chosen for the logarithm. We will explore two approaches as set out in the examiner guide.

Approach 1: Using base 5

Take the $\log_5$ of both sides:

$\log_5 5^{2x} = \log_5 27$

Since $27 = 3^3$, we can rewrite the right hand side:

$\log_5 5^{2x} = \log_5 3^3$

Using the power rule to bring down the exponents:

$2x \log_5 5 = 3 \log_5 3$

As we know $\log_5 5 = 1$, this simplifies to:

$2x = 3 \log_5 3$

Dividing by 2 gives the exact solution:

$x = \frac{3}{2} \log_5 3$

Approach 2: Using base 3

Take the $\log_3$ of both sides:

$\log_3 5^{2x} = \log_3 27$

Since $27 = 3^3$ and $\log_3 3^3 = 3$, we have:

$2x \log_3 5 = 3$

Rearranging for $x$:

$x = \frac{3}{2 \log_3 5}$

Both forms are mathematically equivalent, as can be verified using the change of base formula, though the change of base formula itself is not explicitly tested in the TMUA.

Equations Reducible to Quadratic Form

Some exponential equations do not immediately look like $a^x = b$ but can be reduced to that form using algebraic manipulation. A common type is the 'hidden quadratic', which takes the form $A(a^x)^2 + B(a^x) + C = 0$.

Consider the equation $25^x - 3 \times 5^x + 2 = 0$. To solve this, we must recognise the relationship between the bases. Since $25 = 5^2$, we can use the laws of indices to write $25^x = (5^2)^x = (5^x)^2$.

Substituting $u = 5^x$, the equation becomes a standard quadratic:

$u^2 - 3u + 2 = 0$

Factorising the quadratic:

$(u - 1)(u - 2) = 0$

This gives two possible values for $u$: $u = 1$ or $u = 2$. Now, we substitute back $5^x$ to find $x$:

1. If $5^x = 1$, then $x = 0$ (since $a^0 = 1$ for any positive $a$).
2. If $5^x = 2$, we take logs to find $x = \log_5 2$.

Prior Algebraic Manipulation

In more complex cases, you may need to use the laws of indices or logarithms to consolidate terms before solving. For example, if an equation involves $a^{x+k}$, you should use the rule $a^{x+k} = a^x \times a^k$ to isolate the $a^x$ term. Always ensure that the bases are made consistent across the equation where possible, as this simplifies the application of logarithms later in the process.