Olivia's Solution
Line 1: $\left(4^{2}\right)^{3 x}=\left(4^{3}\right)^{x-2}$
Line 2: $4^{5 x}=4^{3 x-6}$
Line 3: $5 x=3 x-6$
Line 4: $2 x=-6$
Line 5: $x=-3$
Lackson's Solution
Line 1: $\log 16^{3 x}=\log 64^{x-2}$
Line 2: $3 x \log 16=x-2 \log 64$
Line 3: $2 \log 64=x-3 x \log 16$
Line 4: $2 \log 64=x(1-3 \log 16)$
Line 5: $x=\frac{2 \log 64}{1-3 \log 16}$
c) Jackson made one error in his work, which led to an incorrect answer. State the line in which his error falls and explain his error. (1 mark)
d) Algebraically solve the equation again using Jackson's method in the correct way.
(2 marks)
Answer
Final Answer: \(\boxed{-2}\)
Step 1 ：The error in Jackson's solution is in Line 2. He incorrectly applied the logarithm property \(\log a^b = b \log a\). The correct application of this property to the equation in Line 1 would result in \(3x \log 16 = (x-2) \log 64\).
Step 2 ：Now, let's solve the equation correctly using Jackson's method. We will start with the equation in Line 1, apply the logarithm property correctly, and then solve for x.
Step 3 ：Set up the equation: \(3x \log 16 = (x - 2) \log 64\)
Step 4 ：Solve the equation for x, which gives the solution \(x = -2\)
Step 5 ：Final Answer: \(\boxed{-2}\)
