Let's go through each equation and solve for $x$ using natural logarithms:

A) $8^x = 8$

Take the natural logarithm of both sides:

$\ln(8^x) = \ln(8)$

Using the logarithmic identity $\ln(a^b) = b \cdot \ln(a)$:

$x \cdot \ln(8) = \ln(8)$

Since $\ln(8) \neq 0$, divide both sides by $\ln(8)$:

$x = 1$

Answer for A: $x = 1$

B) $8e^x = 5$

Divide both sides by 8:

$e^x = \frac{5}{8}$

Now, take the natural logarithm of both sides:

$x = \ln\left(\frac{5}{8}\right)$

This is the exact value of $x$. If you need an approximate value:

$x \approx \ln(0.625) \approx -0.470$

Answer for B: $x \approx -0.470$

C) $8 \cdot 5^x = 8.5e^{-3x}$

First, divide both sides by 8:

$5^x = \frac{8.5}{8}e^{-3x}$

Simplify the fraction:

$5^x = 1.0625e^{-3x}$

Take the natural logarithm of both sides:

$\ln(5^x) = \ln(1.0625e^{-3x})$

Applying the logarithmic identity $\ln(a \cdot b) = \ln(a) + \ln(b)$:

$x \ln(5) = \ln(1.0625) + \ln(e^{-3x})$

$x \ln(5) = \ln(1.0625) - 3x$

Now, collect the terms involving $x$:

$x \ln(5) + 3x = \ln(1.0625)$

Factor out $x$:

$x(\ln(5) + 3) = \ln(1.0625)$

Solve for $x$:

$x = \frac{\ln(1.0625)}{\ln(5) + 3}$

Using a calculator to approximate:

$x \approx \frac{\ln(1.0625)}{\ln(5) + 3} \approx \frac{0.0606}{1.6094 + 3} \approx \frac{0.0606}{4.6094} \approx 0.013$

Answer for C: $x \approx 0.013$

Would you like further details or explanations on any part?

Related Questions:

1. How can the logarithmic properties help in simplifying equations?
2. What are common logarithmic identities used in solving exponential equations?
3. How does the natural logarithm differ from common logarithms (base 10)?
4. Can you derive a general formula for solving $e^{ax} = b$?
5. What happens when the logarithmic argument is negative or zero?

Tip: When solving exponential equations, remember to isolate the exponential part before applying the logarithm.