Question

a student solved \\(\log_{4}(2x - 12) = 3\\), as shown.
step 1: \\(2x - 12 = 3^{4}\\)
step 2: \\(2x - 12 = 81\\)
step 4: \\(2x = 93\\)
step 5: \\(x = 46.5\\)
explain the error and find the correct solution.

Explanation:

⚡ Using what you learned: Logarithms and Exponential Equations

Step 1: Identify the student's error

The student started with the equation:

$$ \log_{4}(2x - 12) = 3 $$

To rewrite a logarithmic equation of the form \( \log_{b}(A) = C \) into its exponential form, we use the definition:

$$ b^C = A $$

In this problem, the base \( b \) is \( 4 \), the argument \( A \) is \( 2x - 12 \), and the exponent \( C \) is \( 3 \).

The correct exponential form is:

$$ 2x - 12 = 4^3 $$

In Step 1, the student incorrectly wrote:

$$ 2x - 12 = 3^4 $$

The student swapped the base and the exponent, raising \( 3 \) to the power of \( 4 \) instead of raising the base \( 4 \) to the power of \( 3 \).

Step 2: Solve the equation correctly

Now, we solve the equation using the correct exponential form:

$$ 2x - 12 = 4^3 $$

First, evaluate \( 4^3 \):

$$ 4^3 = 4 \times 4 \times 4 = 64 $$

Substitute this back into the equation:

$$ 2x - 12 = 64 $$

Add \( 12 \) to both sides:

$$ 2x = 76 $$

Divide by \( 2 \):

$$ x = 38 $$

Answer:

Error: In Step 1, the student incorrectly converted the logarithmic equation to exponential form. They swapped the base and the exponent, writing \( 3^4 \) instead of \( 4^3 \).

Correct Solution:

$$ \log_{4}(2x - 12) = 3 $$

$$ 2x - 12 = 4^3 $$

$$ 2x - 12 = 64 $$

$$ 2x = 76 $$

$$ x = 38 $$