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Let's solve the problem based on the function \( f(x) = 3^{2x} - 28(3^x) + 27 \).

### Part a) Express \( f(x) \) in the form \( (3^x - a)(3^x - b) \):

We are asked to rewrite the function in the form:
\[
f(x) = (3^x - a)(3^x - b)
\]
This is a quadratic form in terms of \( 3^x \). Let \( y = 3^x \), so the function becomes:
\[
f(y) = y^2 - 28y + 27
\]

Now, we need to factorize this quadratic expression.

We look for two numbers that multiply to 27 and add up to -28. These numbers are -1 and -27.

Thus, the factorization is:
\[
f(y) = (y - 1)(y - 27)
\]

Substituting \( y = 3^x \) back into the equation, we have:
\[
f(x) = (3^x - 1)(3^x - 27)
\]

### Part b) Find the two roots of \( f(x) \):

To find the roots, we set \( f(x) = 0 \), i.e.:
\[
(3^x - 1)(3^x - 27) = 0
\]

This gives us two cases:
1. \( 3^x - 1 = 0 \Rightarrow 3^x = 1 \Rightarrow x = 0 \)
2. \( 3^x - 27 = 0 \Rightarrow 3^x = 27 \Rightarrow x = 3 \) (since \( 27 = 3^3 \))

Thus, the two roots are:
\[
x = 0 \quad \text{and} \quad x = 3
\]

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Would you like more details on any step? Here are some related questions to explore:

1. How do you solve quadratic equations by factorization?
2. What are the general methods to solve exponential equations?
3. Can every quadratic expression be factored into real constants?
4. How does substitution help in simplifying exponential equations?
5. What is the relationship between logarithms and exponentials when solving for \( x \)?

**Tip:** In exponential equations, rewriting the terms in terms of the same base (like \( 3^x \)) can simplify the problem greatly!

The function f(x) is defined as f(x) = 3^{2x} - 28(3^x) + 27. Write f(x) in the form (3^x - a)(3^x - b), where a and b are real constants. Hence find the two roots of f(x).

Learn how to rewrite and solve the exponential function f(x) = 3^{2x} - 28(3^x) + 27 in quadratic form, then find its roots. This problem uses algebraic techniques like factorization and solving exponential equations, ideal for students in Grades 10-12.

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