Question

explain the error in the work shown. find the correct answer.

\\\frac{1}{64} = 16^{2a}\\
\\4^{-3} = (2^4)^{2a}\\
\\4^{-3} = 2^{8a}\\
\\-3 = 8a\\
\\-\frac{3}{8} = a\\

Explanation:

⚡ Using what you learned: Solving Exponential Equations

Step 1: Identify the error in the work

The original problem is:

$$ \frac{1}{64} = 16^{2x} $$

In the second line of the shown work, the student wrote:

$$ 4^{-3} = (2^4)^{2x} $$

While \( \frac{1}{64} = 4^{-3} \) is correct, the bases on both sides must be the same to equate the exponents. The student expressed the left side with a base of \( 4 \) and the right side with a base of \( 2 \) (since \( 16 = 2^4 \)).

To solve correctly, both sides should be written with the same base, such as base \( 4 \) or base \( 2 \).

Step 2: Solve using a common base of 4

Express both sides with base \( 4 \):

$$ \frac{1}{64} = 4^{-3} $$

$$ 16 = 4^2 $$

Substitute these into the equation:

$$ 4^{-3} = (4^2)^{2x} $$

$$ 4^{-3} = 4^{4x} $$

Step 3: Solve for x

Equate the exponents:

$$ -3 = 4x $$

$$ x = -\frac{3}{4} $$

Answer:

Error: The student did not use the same base for both sides of the equation. They wrote the left side with a base of \( 4 \) (\( 4^{-3} \)) and the right side with a base of \( 2 \) (\( 2^4 \)).

Correct Answer:

$$ x = -\frac{3}{4} \text{ (or } -0.75\text{)} $$