Log6(-5r-2)=log6(r-4) Could someone help? I would be very grateful. And can you show the work?

[tex]\log_6(-5r - 2) = \log_6(r-4)[/tex]We can drop the logarithm base-6 because both sides have the same thing. In order for the two logaritms to be equal, the inside parts of the logarithm must be equal. Logarithm is a one-to-one function.[tex]\log_6(-5r - 2) = \log_6(r-4) \Rightarrow -5r - 2 = r- 4 \Rightarrow -6r = - 2 \Rightarrow r = 1/3[/tex]However, this solution does not work. If we try to use r = 1/3 into [tex]\log_6(r-4)[/tex] it yield [tex]\log_6(1/3-4) = \log_6(-11/3)[/tex]. There is no real solution because logarithm cannot be negative.For your new question,[tex]\displaystyle\log_5(w)+\log_5(u/3)-\log_5(v/3) \\= \log_5\left( w \cdot \frac{u}{3} \div \frac{v}{3}\right) \\= \log_5\left( w \cdot \frac{u}{3} \cdot \frac{3}{v}\right) \\= \log_5\left( \frac{wu}{v}\right)[/tex]