Solve for $$ x $$ $$ \log _{6} x-\log _{6}(x+2)=\log _{6}\left(\frac{1}{6}\right) $$ $$ \square $$

Question

UpStudy AI · Accepted Answer

1. Use the logarithm subtraction rule:
   $$
   \log_{6} x - \log_{6}(x+2) = \log_{6}\left(\frac{x}{x+2}\right)
   $$

2. Equate the arguments of the logarithms:
   $$
   \frac{x}{x+2} = \frac{1}{6}
   $$

3. Cross multiply:
   $$
   6x = x+2
   $$

4. Solve for $$ x $$:
   $$
   6x - x = 2 \quad \Rightarrow \quad 5x = 2 \quad \Rightarrow \quad x = \frac{2}{5}
   $$

5. Verify the domain conditions:
   - $$ x > 0 $$ and $$ x+2 > 0 $$, which are both satisfied by $$ x = \frac{2}{5} $$.

Thus, the solution is:
$$
\boxed{\frac{2}{5}}
$$
$$ x = \frac{2}{5} $$

Solve for \( x \) \( \log _{6} x-\log _{6}(x+2)=\log _{6}\left(\frac{1}{6}\right) \) \( \square \)