We are given the equation:

\[
5^{x+10} = 4^{-3x}
\]

To solve for \( x \), we take the natural logarithm (or logarithms of any base) on both sides to make use of the logarithmic property that \( \log(a^b) = b\log(a) \).

### Step 1: Take the natural logarithm of both sides:
\[
\ln(5^{x+10}) = \ln(4^{-3x})
\]

### Step 2: Apply the logarithmic power rule:
\[
(x+10)\ln(5) = (-3x)\ln(4)
\]

### Step 3: Expand both sides:
\[
x\ln(5) + 10\ln(5) = -3x\ln(4)
\]

### Step 4: Group terms involving \( x \) on one side:
\[
x\ln(5) + 3x\ln(4) = -10\ln(5)
\]

### Step 5: Factor out \( x \):
\[
x(\ln(5) + 3\ln(4)) = -10\ln(5)
\]

### Step 6: Solve for \( x \):
\[
x = \frac{-10\ln(5)}{\ln(5) + 3\ln(4)}
\]

Now, we can simplify the expression further by calculating the logarithms. Let's compute the approximate value of \( x \).

The approximate solution for \( x \) is:

\[
x \approx -2.79
\]

Would you like further details on any of the steps or have any questions?

Here are 5 related questions you might find interesting:
1. How do logarithmic properties help in solving exponential equations?
2. Can you solve for \( x \) if the equation had different bases, such as \( 7^{x+2} = 3^{-4x} \)?
3. What happens if both sides of the equation involve the same base, e.g., \( 2^{x+5} = 2^{-3x} \)?
4. How would you solve an equation like \( 10^{2x} = 25^{x-1} \)?
5. How do we solve exponential equations with logarithms of different bases, like \( \log_b(x) \) versus \( \log_c(x) \)?

**Tip:** Always remember to use logarithmic properties like the power rule when solving exponential equations!

Learn how to solve the exponential equation 5^(x+10) = 4^(-3x) by applying logarithmic properties and solving for x. This solution walks through each step, using the logarithmic power rule and properties of exponents. Suitable for students in Grades 10-12.

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