Given that $\log_3 x = a$, find in terms of $a$, (a) $\log_3 (9x)$ (b) $\log_1 \left( \frac{x^2}{81} \right)$, giving each answer in its simplest form - Edexcel - A-Level Maths Pure - Question 7 - 2013 - Paper 5

For $\log_1 \left( \frac{x^2}{81} \right)$, we can again use the properties of logarithms:

$\log_1 \left( \frac{x^2}{81} \right) = \log_1 (x^2) - \log_1 (81).$
Using the power rule for logarithms, $\log_b (m^n) = n \log_b m$, we get:
$\log_1 (x^2) = 2 \log_1 x.$
Since we also have that $\log_1 81 = 4$ (because $81 = 3^4$ and we can convert it to base 3), the expression simplifies further:
$\log_1 \left( \frac{x^2}{81} \right) = 2\log_1 x - 4.$
Substituting $\log_1 x = \log_3 x = a$, we arrive at:
$= 2a - 4.$
The final answer is $2a - 4$.

To solve the equation:
$\log_3 (9x) + \log_3 \left( \frac{x^2}{81} \right) = 3,$
we can substitute the expressions we found in parts (a) and (b):
$\left(2 + a\right) + \left(2a - 4\right) = 3.$
Combining the terms,
$2 + a + 2a - 4 = 3$
which simplifies to:
$3a - 2 = 3.$
Now, solving for $a$:
$3a = 5 \Rightarrow a = \frac{5}{3}.$
Substituting back to find $x$:
Using $\log_3 x = a$, we have:
$\log_3 x = \frac{5}{3}.$
This implies that:
$x = 3^{\frac{5}{3}} = \sqrt[3]{3^5} = \sqrt[3]{243}.$
Calculating this gives approximately $2.498$.
Thus, the answer is $x \approx 2.498$ (to 4 significant figures).