Question

7
Solve the logarithmic equation log8 (6x-2) = log8 (10x-7)
x=

Answer 1

To solve the equation \( \log_8 (6x - 2) = \log_8 (10x - 7) \), we can use the property of logarithms that states if \( \log_b (A) = \log_b (B) \), then \( A = B \), as long as \( A > 0 \) and \( B > 0 \).

This means we can set the arguments of the logarithms equal to each other:

\[
6x - 2 = 10x - 7
\]

Now, let's solve for \( x \):

1. Rearranging the equation:
\[
6x - 10x = -7 + 2
\]

2. This simplifies to:
\[
-4x = -5
\]

3. Dividing both sides by -4 gives:
\[
x = \frac{5}{4}
\]

Now we need to ensure that the solutions do not result in logarithms of non-positive numbers. We will check the conditions:

1. For \( 6x - 2 > 0 \):
\[
6x > 2 \quad \Rightarrow \quad x > \frac{1}{3}
\]

2. For \( 10x - 7 > 0 \):
\[
10x > 7 \quad \Rightarrow \quad x > \frac{7}{10}
\]

Now, check the value \( x = \frac{5}{4} \):

- Since \( \frac{5}{4} = 1.25 > \frac{7}{10} = 0.7 \) and \( \frac{5}{4} > \frac{1}{3} \approx 0.333 \), both conditions are satisfied.

Thus, the solution is:

\[
\boxed{\frac{5}{4}}
\]