We are given the equation:

$5^x + 5^{x+2} + 5^{x+5} = 3^x + 3^{x+3}$

Step 1: Factor the terms with the same base.

For the left-hand side involving powers of 5, we can factor out the common term $5^x$:

$5^x + 5^{x+2} + 5^{x+5} = 5^x (1 + 5^2 + 5^5)$

Calculate $5^2$ and $5^5$:

$5^2 = 25, \quad 5^5 = 3125$

So the expression becomes:

$5^x (1 + 25 + 3125) = 5^x \cdot 3151$

For the right-hand side involving powers of 3, we can factor out $3^x$:

$3^x + 3^{x+3} = 3^x (1 + 3^3)$

Calculate $3^3$:

$3^3 = 27$

So the expression becomes:

$3^x (1 + 27) = 3^x \cdot 28$

Step 2: Set up the simplified equation.

Now we substitute the factored forms back into the original equation:

$5^x \cdot 3151 = 3^x \cdot 28$

Step 3: Take logarithms.

To solve for $x$, we can take the logarithm of both sides (logarithms of any base can be used, but let's use natural logarithms, $\ln$):

$\ln(5^x \cdot 3151) = \ln(3^x \cdot 28)$

Using the properties of logarithms ($\ln(ab) = \ln(a) + \ln(b)$ and $\ln(a^b) = b\ln(a)$):

$x \ln(5) + \ln(3151) = x \ln(3) + \ln(28)$

Step 4: Solve for $x$.

Rearrange the terms to isolate $x$:

$x \ln(5) - x \ln(3) = \ln(28) - \ln(3151)$

Factor out $x$ on the left-hand side:

$x (\ln(5) - \ln(3)) = \ln\left(\frac{28}{3151}\right)$

Now solve for $x$:

$x = \frac{\ln\left(\frac{28}{3151}\right)}{\ln(5) - \ln(3)}$

Step 5: Numerical approximation.

We can calculate the logarithms and the fraction:

$x \approx \frac{\ln\left( \frac{28}{3151} \right)}{\ln(5) - \ln(3)}$

First, compute $\frac{28}{3151} \approx 0.00889$, and then apply the natural logarithm:

$\ln(0.00889) \approx -4.721$

Now, calculate $\ln(5) \approx 1.609$ and $\ln(3) \approx 1.099$:

$\ln(5) - \ln(3) \approx 1.609 - 1.099 = 0.510$

Finally:

$x \approx \frac{-4.721}{0.510} \approx -9.26$

Thus, $x \approx -9.26$.

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Would you like more details on any step or further clarifications?

Here are 5 related questions to expand your understanding:

1. How do logarithmic properties simplify exponential equations?
2. What other techniques can be used to solve exponential equations without logarithms?
3. How do changes in the base of an exponent affect the overall function?
4. What are the key steps in factoring expressions with exponents?
5. Why is natural logarithm commonly used in solving equations involving exponents?

Tip: Always check if the equation can be simplified by factoring before applying logarithms!