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Given that $y = 3x^2$, (a) show that $ ext{log}_3 y = 1 + 2 ext{log}_3 x$ - Edexcel - A-Level Maths Pure - Question 6 - 2012 - Paper 4
Question 6
Given that $y = 3x^2$, (a) show that $ ext{log}_3 y = 1 + 2 ext{log}_3 x$. (b) Hence, or otherwise, solve the equation $1 + 2 ext{log}_3 x = ext{log}_3 (28x... show full transcript
Worked Solution & Example Answer:Given that $y = 3x^2$, (a) show that $ ext{log}_3 y = 1 + 2 ext{log}_3 x$ - Edexcel - A-Level Maths Pure - Question 6 - 2012 - Paper 4
Step 1
show that log3 y = 1 + 2 log3 x
Answer
To show that , we start by substituting into the logarithmic equation:
Using the properties of logarithms, this can be split as follows:
Knowing that and applying the power rule of logarithms (), we have:
Thus, we can rewrite our equation as:
This confirms the required result.
Step 2
Hence, or otherwise, solve the equation 1 + 2 log3 x = log3 (28x - 9)
Answer
Given the equation:
We start by substituting using our previous result:
ightarrow 2 ext{log}_3 x = ext{log}_3 (3x^2) - 1$$ Then, we can rewrite the left side: $$ ext{log}_3 (3x^2) - 1 = ext{log}_3 (28x - 9)$$ Using the property of logarithms which states $ ext{log}_a b - ext{log}_a c = ext{log}_a rac{b}{c}$, we write: $$ ext{log}_3 rac{3x^2}{3} = ext{log}_3 (28x - 9)$$ Therefore: $$ ext{log}_3 (x^2) = ext{log}_3 (28x - 9)$$ Next, we exponentiate both sides to eliminate the logarithm: $$x^2 = 28x - 9$$ Rearranging the equation results in: $$x^2 - 28x + 9 = 0$$ This can be solved using the quadratic formula: $$x = rac{-b ightarrow ext{(from } ax^2 + bx + c = 0 ext{)}}{ ext{2a}}$$ Substituting $a = 1, b = -28, c = 9$ gives: $$x = rac{28 ightarrow ext{ (since } -(-28) = 28)}{2} ightarrow ext{including the sign and assigning properly}$$ $$x = 14 ightarrow ext{ continue to solve.}$$ Finally, substituting back to check the equations. Solving further yields $x = 9$.