The equation of a polynomial is $\text{f}(x) = 2x^3 - 7x^2 + 9x - 3$.

1. Find a linear factor of the polynomial using the Factor Theorem.

Let $\text{f}(x) = 2x^3 - 7x^2 + 9x - 3$.
Try $x = 1$:
$\text{f}(1) = 2(1)^3 - 7(1)^2 + 9(1) - 3 = 2 - 7 + 9 - 3 = 0$.
Since $\text{f}(1) = 0$, $(x - 1)$ is a factor.

2. Divide the polynomial by the linear factor to find the quadratic factor.

Using long division or comparing coefficients:
$2x^3 - 7x^2 + 9x - 3 = (x - 1)(ax^2 + bx + c)$
Comparing $x^3$ coefficients: $a = 2$
Comparing constant terms: $-c = -3 \implies c = 3$
Comparing $x^2$ coefficients: $b - a = -7 \implies b - 2 = -7 \implies b = -5$
So, $\text{f}(x) = (x - 1)(2x^2 - 5x + 3)$.

3. Factorise the quadratic factor completely.

$2x^2 - 5x + 3 = (2x - 3)(x - 1)$
Therefore, $\text{f}(x) = (x - 1)(2x - 3)(x - 1) = (x - 1)^2(2x - 3)$.

1. Use the factorised form of $\text{f}(x)$ from part (i) to set up the equation

From part (i), $\text{f}(x) = (2x - 1)(x^2 - 3x + 3)$.
For $\text{f}(x) = 0$, we have:
$(2x - 1)(x^2 - 3x + 3) = 0$

2. Identify the real root from the linear factor

$2x - 1 = 0 \implies x = \frac{1}{2}$
This gives one real root.

3. Analyse the quadratic factor using the discriminant

For the quadratic factor $x^2 - 3x + 3 = 0$, we check the discriminant $\Delta$:
$\Delta = b^2 - 4ac = (-3)^2 - 4(1)(3)$
$\Delta = 9 - 12 = -3$
Since $\Delta < 0$, the quadratic equation $x^2 - 3x + 3 = 0$ has no real roots.

4. State the final conclusion about the nature of the roots

Therefore, the equation $\text{f}(x) = 0$ has exactly one real root.

1. Rewrite the given equation to reveal the polynomial structure from part (i)

$2(3^y)^3 - 7(3^y)^2 + 3^2(3^y) - 3 = 0$
$2(3^y)^3 - 7(3^y)^2 + 9(3^y) - 3 = 0$

2. Substitute $x = 3^y$ and use the roots of $f(x) = 0$ from earlier parts

Let $x = 3^y$. The equation becomes $f(x) = 0$, which is $2x^3 - 7x^2 + 9x - 3 = 0$.
From part (i), $f(x) = (x-1)^2(2x-3)$.
So the roots are $x = 1$ and $x = \frac{3}{2}$.
Thus, $3^y = 1$ or $3^y = 1.5$.

3. Solve for $y$ using logarithms

For $3^y = 1$:
$y = 0$

For $3^y = 1.5$:
$\ln(3^y) = \ln(1.5)$
$y \ln 3 = \ln 1.5$
$y = \frac{\ln 1.5}{\ln 3} \approx 0.36907$
$y \approx 0.369 \text{ (to 3 s.f.)}$