**Understanding Exponential Equations**

An exponential equation is a mathematical expression that describes exponential relationships, where a variable is positioned in the exponent. These equations typically take the form:

a^{(bx)} = c

Here:

**a**: Base of the exponent, which is a positive real number.**b**: Coefficient of x in the exponent.**x**: The variable situated in the exponent.**c**: A positive real number on the other side of the equation.

Exponential equations are pivotal in various scientific and financial calculations, providing a mathematical framework to describe exponential growth and decay. To fully understand these equations, it's helpful to grasp the basics of geometric sequences.

**Solving Basic Exponential Equations**

**Method 1: Equating Bases and Exponents**

When both sides of the equation have identical bases, we can equate the exponents and solve for the variable.

#### Example 1:

Consider the equation 2^{(3x)} = 2^{6}.

Since the bases are identical, we can set the exponents equal to each other:

3x = 6

Solving for x, we get:

x = 2

**Method 2: Employing Logarithms**

When the bases are not identical, logarithms can be used to solve the equation. For further insight into logarithmic methods, see logarithmic equations.

#### Example 2:

Consider the equation 3^{(2x)} = 9.

First, express all numbers using the same base:

3^{(2x)} = 3^{2}

Now, equate the exponents and solve for x:

2x = 2

x = 1

**Tackling Complex Exponential Equations**

Complex exponential equations may involve additional algebraic manipulation or employing advanced logarithm properties. Understanding the domain and range basics can also be beneficial when solving these equations.

**Method 1: Utilising Logarithm Properties**

Leverage the properties of logarithms to simplify and solve the equation.

#### Example 3:

Consider the equation 5^{(2x+1)} = 25.

Express 25 with the same base as the left side:

5^{(2x+1)} = 5^{2}

Equate the exponents:

2x + 1 = 2

x = 0.5

**Method 2: Algebraic Manipulation**

In some instances, combining like terms or using algebraic manipulation before applying logarithms is necessary. A solid understanding of quadratic functions can be useful for this method.

#### Example 4:

Consider the equation 2^{(x+3)} = 8 * 2^{x}.

Express 8 with the same base:

2^{(x+3)} = 2^{3} * 2^{x}

Use the exponent addition rule:

2^{(x+3)} = 2^{(x+3)}

Since the bases and exponents are equal, the equation is true for all x.

**Real-world Applications of Exponential Equations**

Exponential equations frequently appear in problems related to exponential growth and decay, finance, and physics, offering a mathematical lens through which we can explore and solve real-world problems. Knowledge of the binomial distribution can enhance understanding of these applications.

**Example 5: Investment Problem**

Suppose you invest £1000 in a bank account that offers an annual interest rate of 5%, compounded annually. How many years will it take for your investment to double?

The formula for compound interest is:

A = P * (1 + r)^{n}

Where:

**A**: Amount after n years**P**: Principal amount (£1000)**r**: Annual interest rate (0.05)**n**: Number of years

We want to find out when A will be £2000:

2000 = 1000 * (1 + 0.05)^{n}

2 = (1.05)^{n}

To solve for n, we can use logarithms:

log(2) = n * log(1.05)

n ≈ 14.21

Therefore, it will take approximately 14 years for the investment to double.

**Recap and Practice**

Exponential equations are crucial for understanding various mathematical and real-world phenomena. The methods outlined above, including using the same base, applying logarithms, and algebraic manipulation, are fundamental in solving these equations. For further study, see the introduction to logarithmic equations.

**Practice Questions**

- Solve the exponential equation: 4
^{(x+2)}= 64 - Solve for x: 7
^{(2x)}= 49 - If £5000 is invested at an annual interest rate of 6%, find how many years it will take for the investment to grow to £8000.

## FAQ

When coefficients are present in front of the exponents, such as 2 * 3^{(x)} = 12, you should first isolate the exponential term by performing algebraic operations. In this case, divide both sides of the equation by 2 to get 3^{(x)} = 6. From here, you can use logarithms to solve for x. Taking the natural logarithm on both sides, we get ln(3^{(x)}) = ln(6). Applying the logarithm power rule, we have x * ln(3) = ln(6). Finally, solve for x by dividing by ln(3), resulting in x = ln(6) / ln(3). This method ensures that the coefficient does not hinder your ability to solve the exponential equation.

Yes, exponential equations can have no solution. For instance, consider the equation 2^{(x)} = -4. Since the result of any non-zero number raised to an exponent is always positive, there is no real number solution for x that will satisfy 2^{(x)} = -4. It's crucial to note that the range of an exponential function with a positive base is (0, ∞), meaning it will never yield a negative value. Therefore, whenever you encounter an exponential equation equating to a negative number, it has no solution in the real number system.

Solving an equation like x^{(2x)} = 32 can be complex due to the presence of the variable in both the base and the exponent. One approach is to take the natural logarithm of both sides: ln(x^{(2x)}) = ln(32). Applying the logarithm power rule, we get 2x * ln(x) = ln(32). At this point, the equation may still be complex to solve algebraically, and graphical or numerical methods might be employed. Graphically, you might observe the behaviour of the function y = 2x * ln(x) - ln(32) and find where it intersects the x-axis. Numerical methods, like the Newton-Raphson method, can refine an approximate root to a more accurate value.

To solve an exponential equation with different bases, such as a^{(x)} = b, you can use logarithms to simplify the equation. Taking the logarithm of both sides of the equation, we get ln(a^{(x)}) = ln(b). Utilising the logarithm power rule (ln(c^{d}) = d * ln(c)), the equation becomes x * ln(a) = ln(b). To isolate x, divide both sides of the equation by ln(a), resulting in x = ln(b) / ln(a). This method allows you to solve for x even when the bases of the exponential expressions are different, providing a systematic approach to handling various exponential equations.

When the variable appears in both the base and the exponent of an equation, such as x^{(x)} = a, logarithms can be particularly useful. To solve x^{(x)} = a, take the natural logarithm (ln) on both sides of the equation. Using the logarithm power rule, which allows us to move the exponent in front of the logarithm, we get x * ln(x) = ln(a). This equation might still be complex to solve algebraically, and in some cases, graphical methods or numerical methods (like the Newton-Raphson method) might be employed to find the value of x. It’s essential to graphically observe the behaviour of the function and determine the approximate value of the root, which can then be refined using numerical methods.

## Practice Questions

The given equation is 3^{(2x + 1)} = 81. To solve for x, we can express 81 as a power of 3 since they have the same base, which gives us 3^{(2x + 1)} = 3^{4}. Since the bases are the same, we can equate the exponents: 2x + 1 = 4. Solving for x, we subtract 1 from both sides of the equation, getting 2x = 3, and then divide by 2 to isolate x, which gives us x = 1.5. To validate the answer, we substitute x = 1.5 back into the original equation: 3^{(2*1.5 + 1)} = 3^{4}, which simplifies to 3^{4} = 3^{4}, confirming that x = 1.5 is correct.

The formula for compound interest is A = P * (1 + r)^{n}, where A is the amount after n years, P is the principal amount (£1000), r is the annual interest rate (0.05), and n is the number of years. We want to find out when A will be £3000: 3000 = 1000 * (1 + 0.05)^{n}. Simplifying, we get 3 = (1.05)^{n}. To solve for n, we can use logarithms: log(3) = n * log(1.05). Solving for n, we get n ≈ log(3) / log(1.05) ≈ 22.52. Therefore, it will take approximately 23 years for the investment to triple, as we always round up in the context of years in real-life situations.

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.