To solve the general exponential equation $a^x = b$, where $a > 0$ and $a \neq 1$, follow these steps:

1. Take the logarithm of both sides:

Apply the logarithm to both sides of the equation. You can use any logarithm, but it's common to use the natural logarithm ($\ln$) or the common logarithm ($\log$). For this explanation, we'll use the natural logarithm.

$\ln(a^x) = \ln(b)$

2. Use the power rule of logarithms:

The power rule states that $\ln(a^x) = x \cdot \ln(a)$. Apply this to the left side:

$x \cdot \ln(a) = \ln(b)$

3. Solve for $x$:

Now, isolate $x$ by dividing both sides by $\ln(a)$:

$x = \frac{\ln(b)}{\ln(a)}$

Example:

Let's say you have $3^x = 81$. To solve it:

1. Take the logarithm of both sides:

   $\ln(3^x) = \ln(81)$

2. Use the power rule:

   $x \cdot \ln(3) = \ln(81)$

3. Solve for $x$:

   $x = \frac{\ln(81)}{\ln(3)}$

   Since $81 = 3^4$, $x = 4$.

Would you like more details or have other questions?

Here are 8 related questions you might find useful:

1. How can you solve $2^x = 50$ using logarithms?
2. What is the difference between using natural logarithms and common logarithms in these equations?
3. How would you solve $5^x = 100$?
4. Can you solve $a^x = b$ if $a$ or $b$ is a negative number?
5. What happens if $a = 1$ in the equation $a^x = b$?
6. How do you solve exponential equations with different bases, like $2^x = 3^y$?
7. How would you solve an equation like $3^{2x} = 81$?
8. What is the method for solving $e^x = 7$, where $e$ is the base of natural logarithms?

Tip: When solving exponential equations, always consider the properties of logarithms and how they can simplify the process.