The problem is to solve the equation:

$3^{2x} - 3^{x+1} = 10$

Step-by-Step Solution:

Step 1: Rewrite $3^{2x}$ in terms of $3^x$

We know that: $3^{2x} = (3^x)^2$

So the equation becomes: $(3^x)^2 - 3 \cdot 3^x = 10$

Step 2: Let $y = 3^x$

Substituting $y$ for $3^x$, the equation becomes: $y^2 - 3y - 10 = 0$

Step 3: Solve the quadratic equation

The equation $y^2 - 3y - 10 = 0$ can be factored as: $(y - 5)(y + 2) = 0$

Thus: $y = 5 \quad \text{or} \quad y = -2$

Step 4: Back-substitute $y = 3^x$

Since $3^x > 0$, we reject $y = -2$. Therefore: $3^x = 5$

Step 5: Solve for $x$

Take the logarithm of both sides: $x \ln(3) = \ln(5)$

Solve for $x$: $x = \frac{\ln(5)}{\ln(3)}$

Final Answer:

$x = \frac{\ln(5)}{\ln(3)} \approx 1.46497$

Would you like a detailed explanation of any step?

Related Questions:

1. What are logarithms, and how are they applied to solve exponential equations?
2. Can you solve the same equation graphically?
3. How does rejecting $y = -2$ affect the solution in exponential equations?
4. What is the general process of converting a quadratic exponential equation into a solvable form?
5. Can this equation have complex solutions?

Tip: To solve exponential equations, substituting variables and working through logarithmic operations is often a clean and effective approach.