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You provided three equations:

1. \( e^{4 - x} = 3 \)  (f)
2. \( 2 \cdot e^{6 - 3z} = 2 \) (g)
3. \( e^{3 - 4a} = 5 \) (h)

Let's solve each one individually.

### f) \( e^{4 - x} = 3 \)
We take the natural logarithm (ln) on both sides to solve for \(x\):

\[
\ln(e^{4 - x}) = \ln(3)
\]

\[
4 - x = \ln(3)
\]

\[
x = 4 - \ln(3)
\]

Thus, \( x = 4 - \ln(3) \).

### g) \( 2 \cdot e^{6 - 3z} = 2 \)

First, divide both sides by 2:

\[
e^{6 - 3z} = 1
\]

Now, take the natural logarithm of both sides:

\[
\ln(e^{6 - 3z}) = \ln(1)
\]

Since \( \ln(1) = 0 \):

\[
6 - 3z = 0
\]

Solve for \(z\):

\[
3z = 6 \quad \Rightarrow \quad z = 2
\]

### h) \( e^{3 - 4a} = 5 \)

Take the natural logarithm of both sides:

\[
\ln(e^{3 - 4a}) = \ln(5)
\]

\[
3 - 4a = \ln(5)
\]

Solve for \(a\):

\[
4a = 3 - \ln(5) \quad \Rightarrow \quad a = \frac{3 - \ln(5)}{4}
\]

Thus, \( a = \frac{3 - \ln(5)}{4} \).

---

Would you like additional details or clarifications? 

Here are 5 related questions to expand on these:

1. How do you simplify natural logarithms involving exponents, like \( e^x \)?
2. How would the solutions change if the base of the exponent wasn't \(e\)?
3. How do you solve logarithmic equations with more complex coefficients?
4. What are the properties of natural logarithms that help in solving such equations?
5. How can these logarithmic solving techniques apply to real-world exponential growth problems?

**Tip**: The natural logarithm function, \( \ln(x) \), is especially useful for solving equations involving \(e^x\) since it "undoes" the exponentiation.

Solve the equations e^(4 - x) = 3, 2 * e^(6 - 3z) = 2, and e^(3 - 4a) = 5.

Learn how to solve exponential equations like e^(4 - x) = 3, 2 * e^(6 - 3z) = 2, and e^(3 - 4a) = 5 using natural logarithms. This solution walks through each step, from applying logarithms to solving for variables x, z, and a. Suitable for students in Grades 10-12.

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