Question

Solve each exponential equation. Express irrational solutions in exact form. $$ 36^{x}-6 \cdot 6^{x}=-9 $$

Short answer

The solution is \(x = \frac{\text{ln}(3)}{\text{ln}(6)}\).

Step-by-step solution

Rewrite the Equation with a Common Base

Notice that 36 can be written as \(6^2\) and rewrite the equation to: \((6^2)^x - 6 \times 6^x = -9\)

Simplify the Exponents

Using the property of exponents \((a^m)^n = a^{mn}\), rewrite \(6^{2x} - 6 \times 6^x = -9\).

Let Substitution

For easier handling, let \(y = 6^x\). The equation now becomes \(y^2 - 6y = -9\).

Rewrite as a Quadratic Equation

Rewrite the equation in standard quadratic form: \(y^2 - 6y + 9 = 0\).

Solve the Quadratic Equation

Factor the quadratic equation: \((y - 3)^2 = 0\), hence \(y = 3\).

Reverse the Substitution

Since \(y = 6^x\), we have \(6^x = 3\).

Solve for x

Taking the natural logarithm on both sides, \(\text{ln}(6^x) = \text{ln}(3)\). Apply the power rule: \(x \text{ln}(6) = \text{ln}(3)\), hence \(x = \frac{\text{ln}(3)}{\text{ln}(6)}\).

Key concepts

Quadratic Equations

A quadratic equation is a second-order polynomial equation in a single variable. It has the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants, and \(a e 0\).
In the given exercise, we encounter a quadratic equation after substituting \(y = 6^x\). The rewritten equation, \(y^2 - 6y + 9 = 0\), fits the standard quadratic form. Quadratic equations can be solved using methods such as:

• Factoring
• Completing the square
• Using the quadratic formula: \(\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

In the example, we factored the equation into \((y - 3)^2 = 0\), which simplifies to \(y = 3\).
Understanding how to rewrite and factor quadratic equations is a valuable skill in algebra and higher mathematics. By solving them, we can find the values of the variable that satisfy the equation.

Natural Logarithms

The natural logarithm, denoted as \(\text{ln}\), is the logarithm to the base \(e\) (where \(e\) is approximately equal to 2.718).
Natural logarithms have properties that make them useful for solving exponential equations.
In the given problem, we use natural logarithms to solve \(6^x = 3\). By taking the natural logarithm on both sides, we can extract \(x\) from the exponent.
The properties of natural logarithms that are useful in such problems include:

• \(\text{ln}(ab) = \text{ln}(a) + \text{ln}(b)\)
• \(\text{ln}\left(\frac{a}{b}\right) = \text{ln}(a) - \text{ln}(b)\)
• \(\text{ln}(a^b) = b \cdot \text{ln}(a)\)

In this exercise, we used the power rule \(\text{ln}(6^x) = x \cdot \text{ln}(6)\) to solve for \(x\). By doing so, \(x = \frac{\text{ln}(3)}{\text{ln}(6)}\) was obtained.
Mastering natural logarithms can make solving exponential equations more straightforward and intuitive.

Exponent Properties

Understanding the properties of exponents is essential for manipulating and solving exponential equations.
The critical properties of exponents used in the given exercise include:

• Product of powers: \(a^m \cdot a^n = a^{m+n}\)
• Power of a power: \((a^m)^n = a^{mn}\)
• Power of a product: \((ab)^n = a^n \cdot b^n\)

Initially, we rewrote \(36\) as \((6^2)\), and hence, the term \((6^2)^x\) was simplified using the power of a power property to \(6^{2x}\).
This rewiring allowed us to create a quadratic equation by setting \(y = 6^x\). By understanding and applying these exponent properties, exponential equations become much easier to manage.
These principles are foundational for more advanced topics in algebra and calculus. Practice using these properties with different bases and exponents to become more proficient in solving similar problems.