In logarithmic equations, where the unknown is an exponent, are fundamental. These equations require a deep understanding of logarithms and exponents, and their interplay is crucial for solving complex mathematical problems.

**Understanding Logarithmic Equations**

Logarithmic equations, where the unknown is in the exponent, are solved by exploiting the relationship between logarithms and exponents. This involves using the inverse nature of logarithms to 'undo' exponential functions and applying logarithmic laws to simplify and solve the equations.

**Key Components:**

**1. Exponential Functions**:

Represented as $a^x$, where $a$ is a positive constant and $x$ is the exponent.

**2. Logarithmic Functions**:

Denoted as $\log_a(x)$, these are the inverse of exponential functions, answering "To what power must $a$ be raised to obtain $x$?".

**3. Logarithmic Laws**:

Critical for simplifying and solving logarithmic equations.

- Product Law: $\log_a(xy) = \log_a(x) + \log_a(y)$
- Quotient Law: $\log_a\left(\frac{x}{y}\right) = \log_a(x) - \log_a(y)$
- Power Law: $\log_a(x^b) = b \log_a(x)$
- Change of Base Formula: $\log_b(x) = \frac{\log_a(x)}{\log_a(b)}.$

**Solving Strategy:**

**1. Isolate** the exponential part of the equation.

**2. Apply Logarithms** on both sides, using a common base like 10 or $e$.

**3. Utilize Logarithm Laws** to simplify the equation into a more solvable form.

**4. Solve for the Variable** using algebraic techniques.

**Examples:**

**Example 1: Solve **25^x < 1

**Convert to Logarithmic Form:**$25$ as $5^2$, so 5^{2x} < 1.**Apply Logarithms:**\log5^{2x} < \log(1).**Simplify Using Logarithm Laws:**2x \log5 < 0 since $\log1 = 0.$**Solve for**$x$**:**As \log5 > 0, x < 0.

**Answer:** x < 0.

**Example 2: Solve **3^{2x} < 5^{3x-1}

**Apply Logarithms:**\log3^{2x} < \log5^{3x-1}.**Use Logarithm Laws:**Simplify to $2x \log3$< $3x-1 \log5.$**Rearrange and Solve:**2x \log(3) - 3x \log(5) < -\log(5), simplifying to x(2\log(3) - 3\log(5)) < -\log(5)

**Answer:** Solve for$x$: x < \frac{-\log(5)}{2\log(3) - 3\log(5)} = 0.6117.

**Example 3: Solve **$2^{3x} = 8$

**Recognize the Equation:**$2^{3x} = 8.$**Simplify the Equation:**8 as $2^3,$ so $2^{3x} = 2^3.$**Equating Exponents:**As bases are the same, $3x = 3.$**Solve for**$x:$

**Answer: **$x = 1.$

Rahil spent ten years working as private tutor, teaching students for GCSEs, A-Levels, and university admissions. During his PhD he published papers on modelling infectious disease epidemics and was a tutor to undergraduate and masters students for mathematics courses.