(b) Solve the equation 3^(2*x*+1)−3^(*x*+2)+6=0  

For part (b), the given equation is:

\[
3^{2x+1} - 3^{x+2} + 6 = 0
\]

Let's solve it step-by-step:

### Step 1: Simplify the Expression

Rewrite the equation to express all terms with the same base:

\[
3^{2x+1} - 3^{x+2} + 6 = 0
\]

We can express \(3^{2x+1}\) as \((3^x)^2 \cdot 3\) and \(3^{x+2}\) as \(3^x \cdot 9\). Let \(z = 3^x\). Then:

\[
3^{2x+1} = 3 \cdot z^2, \quad 3^{x+2} = 9z
\]

Substituting these into the equation gives:

\[
3z^2 - 9z + 6 = 0
\]

### Step 2: Solve the Quadratic Equation

This is a quadratic equation in terms of \(z\). Divide everything by 3 to simplify:

\[
z^2 - 3z + 2 = 0
\]

The quadratic formula is:

\[
z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

where \(a = 1\), \(b = -3\), and \(c = 2\). Substituting these values in:

\[
z = \frac{-(-3) \pm \sqrt{(-3)^2 - 4 \cdot 1 \cdot 2}}{2 \cdot 1}
\]
\[
z = \frac{3 \pm \sqrt{9 - 8}}{2}
\]
\[
z = \frac{3 \pm \sqrt{1}}{2}
\]
\[
z = \frac{3 \pm 1}{2}
\]

So, we have two possible solutions for \(z\):

\[
z_1 = \frac{3 + 1}{2} = 2, \quad z_2 = \frac{3 - 1}{2} = 1
\]

### Step 3: Find \(x\) from \(z\)

Recall that \(z = 3^x\). Thus, we have two equations to solve:

1. \(3^x = 2\)
2. \(3^x = 1\)

For the first equation:

\[
x = \log_3 2
\]

For the second equation:

\[
3^x = 1 \implies x = 0
\]

### Step 4: Provide Solutions in Required Format

The solutions for \(x\) are:

\[
x = 0, \quad x = \log_3 2
\]

To express \(\log_3 2\) in 3 significant figures, we can use the change of base formula:

\[
\log_3 2 = \frac{\log 2}{\log 3} \approx 0.631
\]

### Final Answer

The solutions to the equation, in ascending order and rounded to three significant figures, are:

\[
x = 0 \quad \text{or} \quad x = 0.631
\]

Would you like further details or have any questions about this solution?

### Follow-Up Questions:
1. How can logarithmic properties help simplify exponential equations?
2. What are some real-life applications of solving exponential equations?
3. How do you derive the change of base formula for logarithms?
4. What is the difference between natural logarithms and common logarithms?
5. How can you solve exponential equations graphically?

**Tip:** When solving exponential equations, substituting a new variable for the exponent can simplify the problem to a basic quadratic form.

Learn how to solve the exponential equation 3^(2x+1) - 3^(x+2) + 6 = 0 step-by-step with a detailed solution. This problem involves exponential and quadratic equations, logarithms, and the quadratic formula. Suitable for high school students.

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