Question

Solve each exponential equation. Express irrational solutions in exact form. $$ 2^{2 x}+2^{x}-12=0 $$

Short answer

The solution is \( x = \frac{\text{ln}(3)}{\text{ln}(2)} \).

Step-by-step solution

Identify the Substitution

First, recognize that the equation is in the form that can be simplified using a substitution. Let us set a new variable, for instance, let us say \(y = 2^x\).

Substitute the Variable

Replace \(2^x\) with \(y\) in the given equation. This gives us the new equation: \(y^2 + y - 12 = 0\).

Solve the Quadratic Equation

Now, solve the quadratic equation \(y^2 + y - 12 = 0\). We can factor this equation: \[ (y + 4)(y - 3) = 0 \]. Therefore, \(y + 4 = 0\) or \(y - 3 = 0\). This gives us the solutions \(y = -4\) and \(y = 3\).

Back Substitute \(y = 2^x\)

Now, revert the substitution \(y = 2^x\). We have two cases to consider: 1. \(2^x = -4\) which is not possible since a power of 2 cannot be negative. 2. \(2^x = 3\).

Solve for \(x\)

We solve \(2^x = 3\) by taking the logarithm of both sides. Using the natural logarithm, we get: \[ \text{ln}(2^x) = \text{ln}(3) \]. Using the properties of logarithms, this simplifies to: \[ x \text{ln}(2) = \text{ln}(3) \]. Finally, solve for \(x\): \[ x = \frac{\text{ln}(3)}{\text{ln}(2)} \].

Express the Solution in Exact Form

The solution is \( x = \frac{\text{ln}(3)}{\text{ln}(2)} \). This is the exact form of the irrational solution.

Key concepts

exponential functions

Exponential functions are important in various fields, including science, finance, and engineering. An exponential function is a function in which the variable is an exponent. For example, in the function \(f(x) = a^x\), 'a' is a constant and 'x' is the variable. Here are a few key points:

• The base 'a' is a positive real number and cannot be equal to 1.
• Exponential growth occurs when 'a' is greater than 1, and exponential decay occurs when 'a' is between 0 and 1.
• These functions have constant growth or decay rates, meaning the rate of change is proportional to the current value.

In the given exercise, we have an exponential equation with the base 2: \(2^{2x} + 2^x - 12 = 0\).
By substituting \(2^x\) with a new variable \(y\), we simplified the problem into a quadratic equation. This substitution makes it easier to solve equations where the exponent is a variable.

logarithms

Logarithms are the inverses of exponential functions. If we have an exponential function \(a^x = y\), the corresponding logarithmic form is \(x = \text{log}_a(y)\). Here are some key points about logarithms:

• Common logarithms use base 10: \(\text{log}_{10}(x)\). Natural logarithms use base 'e' (approximately 2.718): \(\text{ln}(x)\).
• Logarithms have properties that simplify calculations, such as \(\text{log}_a(m \times n) = \text{log}_a(m) + \text{log}_a(n)\) and \(\text{log}_a(\frac{m}{n}) = \text{log}_a(m) - \text{log}_a(n)\).

In the solution to our problem, we used logarithms to solve for 'x'. Specifically, we took the natural logarithm of both sides of the equation \(2^x = 3\), resulting in:

\[ \text{ln}(2^x) = \text{ln}(3) \
\text{Using the property: } x \text{ln}(2) = \text{ln}(3) \
x = \frac{\text{ln}(3)}{\text{ln}(2)} \].
This exact form provides the irrational solution.

quadratic equations

Quadratic equations are polynomials of degree 2, generally in the form \(ax^2 + bx + c = 0\). They can be solved using various methods: factoring, completing the square, or using the quadratic formula. Key points include:

• The quadratic formula is \(x = \frac{-b \text{±} \text{√}(b^2 - 4ac)}{2a}\).
• Factoring is often the simplest method if the quadratic can be written as a product of binomials.
• The discriminant, \(b^2 - 4ac\), determines the nature of the roots: If positive, there are 2 real roots; if zero, 1 real root; if negative, roots are complex.

In the exercise, after substituting \(y = 2^x\), we got the quadratic equation \(y^2 + y - 12 = 0\).
We factored it as \((y + 4)(y - 3) = 0\), leading to solutions \(y = -4\) (not valid for \(2^x\)) and \(y = 3\).
We then reverted the substitution and solved \(2^x = 3\) using logarithms.
Thus, the use of quadratic techniques simplified our process.