Given that a and b are positive constants, solve the simultaneous equations a = 3b, log₃ a + log₂ b = 2 - Edexcel - A-Level Maths Pure - Question 7 - 2008 - Paper 2

Next, substitute 'a' in the second equation:

log₃ (3b) + log₂ b = 2.

Using the property of logarithms, we can rewrite this:

log₃ (3b) = log₃ 3 + log₃ b = 1 + log₃ b.

Thus, the equation becomes:

1 + log₃ b + log₂ b = 2.

Subtracting 1 from both sides gives:

log₃ b + log₂ b = 1.

To find 'b', let:

let x = log₃ b. Then, using the change of base formula, we have:

log₂ b = log₂ (3^x) = x * log₂ 3.

Therefore, the equation becomes:

x + x * log₂ 3 = 1.

Factoring out 'x' results in:

x(1 + log₂ 3) = 1.

Thus, x = \frac{1}{1 + log₂ 3}.

Now, substituting back to find 'b':

b = 3^x = 3^{\frac{1}{1 + log₂ 3}}.

Now substituting the value of 'b' back into the expression for 'a':

a = 3b = 3 * 3^{\frac{1}{1 + log₂ 3}} = 3^{\frac{1 + 1 + log₂ 3}{1 + log₂ 3}}.

Finally, both 'a' and 'b' can now be stated as exact numbers:

a = 3^{\frac{2 + log₂ 3}{1 + log₂ 3}}, b = 3^{\frac{1}{1 + log₂ 3}}.