Solve for x: log6 (2x − 4) + log6 (4) = log6 40

Question

ScanSolve AI Tutor · Accepted Answer

yes, the method shown in the image is correct, and the steps lead to the right answer. There's just a minor notation point in the third line.

Let's go through the steps clearly:

Given the equation:

\log_6 (2x - 4) + \log_6 (4) = \log_6 40

Apply the product rule of logarithms, $\log_b A + \log_b B = \log_b (AB)$ , to the left side of the equation.
$\log_6 ( (2x - 4) \cdot 4 ) = \log_6 40$
Distribute the 4 into the term $(2x - 4)$ .
$\log_6 (8x - 16) = \log_6 40$
(Note: The image showed $\log_6 8x - 16 = \log_6 40$ , which could be misinterpreted as $\log_6(8x)$ minus 16. The correct interpretation, as used in the subsequent steps, is that $8x-16$ is the argument of the logarithm, so it should be enclosed in parentheses.)
Since the logarithms on both sides have the same base and are equal, their arguments must be equal.
$8x - 16 = 40$
Add 16 to both sides of the equation.
$8x = 40 + 16$ $8x = 56$
Divide both sides by 8 to solve for $x$ .
$x = \frac{56}{8}$ $x = 7$
Check the validity of the solution.
The argument of a logarithm must be positive.
For $x = 7$ :
$2x - 4 = 2(7) - 4 = 14 - 4 = 10$ . Since $10 > 0$ , the logarithm $\log_6(10)$ is defined.
The other arguments (4 and 40) are already positive.
Thus, $x=7$ is a valid solution.

The final answer is $\boxed{7}$ .

Solve for x: log6 (2x - 4) + log6 (4) = log6 40