Question

Solve each exponential equation. Express irrational solutions in exact form. $$ 9^{x}-3^{x+1}+1=0 $$

Short answer

x = \frac{\ln \left( \frac{3 + \sqrt{5}}{2} \right)}{\ln (3)} or \frac{\ln \left( \frac{3 - \sqrt{5}}{2} \right)}{\ln (3)}

Step-by-step solution

Rewrite Terms with Common Base

Notice that 9 can be written as a power of 3. Rewrite the equation by expressing 9 as 3^2: \[ 9^{x} - 3^{x+1} + 1 = 0 \] can be rewritten as \[ (3^2)^{x} - 3^{x+1} + 1 = 0 \] which simplifies to: \[ 3^{2x} - 3^{x+1} + 1 = 0 \]

Use Substitution

Let \( y = 3^x \). Then, the equation becomes: \[ y^2 - 3y + 1 = 0 \]

Solve the Quadratic Equation

Solve the quadratic equation \( y^2 - 3y + 1 = 0 \) using the quadratic formula: \[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a = 1 \), \( b = -3 \), and \( c = 1 \). This gives: \[ y = \frac{3 \pm \sqrt{9 - 4}}{2} \] which simplifies to: \[ y = \frac{3 \pm \sqrt{5}}{2} \]

Replace Back

Recall that \( y = 3^x \). So, we have: \[ 3^x = \frac{3 + \sqrt{5}}{2} \] and \[ 3^x = \frac{3 - \sqrt{5}}{2} \]

Solve for x

Take the natural logarithm of both sides for each equation. For the first equation: \[ x \ln(3) = \ln \left( \frac{3 + \sqrt{5}}{2} \right) \] which simplifies to: \[ x = \frac{\ln \left( \frac{3 + \sqrt{5}}{2} \right)}{\ln (3)} \] For the second equation: \[ x \ln(3) = \ln \left( \frac{3 - \sqrt{5}}{2} \right) \] which simplifies to: \[ x = \frac{\ln \left( \frac{3 - \sqrt{5}}{2} \right)}{\ln (3)} \]

Key concepts

Solving Exponential Equations

Exponential equations involve variables in the exponent, and solving them typically requires just a few strategic steps. Let's consider an example: \ \(9^{x}-3^{x+1}+1=0\) \
The key is to rewrite terms so that they share a common base. For instance, 9 can be expressed as \(3^2\). This allows us to manipulate the equation more easily: \ \(9^x - 3^{x+1} + 1 = 0\) \ becomes \ \((3^2)^x - 3^{x+1} + 1 = 0\).
This simplifies further to \(3^{2x} - 3^{x+1} + 1 = 0\). Using substitution, you can turn this exponential equation into a simpler form. This is a critical part of solving such problems.

Using Substitution in Algebra

Substitution is a powerful technique in algebra to simplify complex expressions. In our previous example, let’s substitute \(y = 3^x\). This transforms the equation: \ \(3^{2x} - 3^{x+1} + 1 = 0\) \ into a quadratic form: \ \(y^2 - 3y + 1 = 0\).
When done correctly, substitution makes an exponential equation much easier to solve. After solving for \(y\), remember to reverse the substitution to find the original variable.

Quadratic Formula

The quadratic formula is an essential tool for solving equations of the form \(ax^2 + bx + c = 0\). Here, \(a = 1\), \(b = -3\), and \(c = 1\). The formula is: \ \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) \ Plugging in our values, we get: \ \(y = \frac{3 \pm \sqrt{9 - 4}}{2} = \frac{3 \pm \sqrt{5}}{2}\).
This provides the solutions \(y = \frac{3 + \- \sqrt{5}}{2}\). These values are instrumental in solving the original exponential equation once the substitution is reversed.

Natural Logarithms

Natural logarithms (ln) are used to undo exponentiation involving the base \(e\). To solve for \(x\) in exponential forms like \(3^x\), we use the natural logarithm. Taking the natural logarithm of both sides helps isolate \(x\). For example: \ Let's solve \(3^x = \frac{3 + \sqrt{5}}{2}\).
Taking the natural logarithm of both sides gives: \ \(x \ln(3) = \ln(\frac{3 + \sqrt{5}}{2})\), which simplifies to: \ \(x = \frac{\ln(\frac{3 + \sqrt{5}}{2})}{\ln(3)}\).
This process is repeated for \(3^x = \frac{3 - \sqrt{5}}{2}\) to find the second value for \(x\).