We are given the equation:

$2^{2x+8} = 3^{x-3}$

To solve for $x$, we can use logarithms. Here’s the step-by-step approach:

Step 1: Take the logarithm of both sides

We can take the natural logarithm (ln) or the logarithm to any base, but we'll use the natural logarithm for convenience:

$\ln\left(2^{2x+8}\right) = \ln\left(3^{x-3}\right)$

Step 2: Apply the power rule of logarithms

The power rule states that $\ln(a^b) = b \ln(a)$. So we can rewrite both sides of the equation:

$(2x+8) \ln(2) = (x-3) \ln(3)$

Step 3: Expand and rearrange terms

Now, distribute the logarithms:

$(2x+8) \ln(2) = (x-3) \ln(3)$

Expanding both sides:

$2x \ln(2) + 8 \ln(2) = x \ln(3) - 3 \ln(3)$

Now, collect all terms involving $x$ on one side of the equation:

$2x \ln(2) - x \ln(3) = -8 \ln(2) - 3 \ln(3)$

Step 4: Factor out $x$

Factor $x$ from the left-hand side:

$x \left( 2 \ln(2) - \ln(3) \right) = -8 \ln(2) - 3 \ln(3)$

Step 5: Solve for $x$

Now, isolate $x$:

$x = \frac{-8 \ln(2) - 3 \ln(3)}{2 \ln(2) - \ln(3)}$

This is the exact solution for $x$. You can plug in the values of $\ln(2)$ and $\ln(3)$ to compute a numerical answer if desired:

• $\ln(2) \approx 0.6931$
• $\ln(3) \approx 1.0986$

Step 6: Final computation (optional)

Substitute the values:

$x = \frac{-8(0.6931) - 3(1.0986)}{2(0.6931) - 1.0986}$

Now calculate:

$x = \frac{-5.5448 - 3.2958}{1.3862 - 1.0986}$

$x = \frac{-8.8406}{0.2876}$

$x \approx -30.7$

Final Answer:

So, $x \approx -30.7$.

Would you like further details on any step? Here are some questions for you:

1. What are the properties of logarithms used in this solution?
2. How can you use logarithms for equations involving exponents?
3. What are some other methods for solving exponential equations?
4. Why do we use natural logarithms in this case instead of logarithms with other bases?
5. How would the solution change if the equation had a different base on each side?

Tip: Always double-check your calculations, especially when dealing with logarithms and exponentials, as they often involve approximation.