154k views
1 vote
Log6(-5r-2)=log6(r-4) Could someone help? I would be very grateful. And can you show the work?

1 Answer

5 votes


\log_6(-5r - 2) = \log_6(r-4)

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.


\log_6(-5r - 2) = \log_6(r-4) \Rightarrow -5r - 2 = r- 4 \Rightarrow -6r = - 2 \Rightarrow r = 1/3

However, this solution does not work. If we try to use r = 1/3 into
\log_6(r-4) it yield
\log_6(1/3-4) = \log_6(-11/3). There is no real solution because logarithm cannot be negative.

For your new question,


\displaystyle\log_5(w)+\log_5(u/3)-\log_5(v/3) \\= \log_5\left( w \cdot (u)/(3) / (v)/(3)\right) \\= \log_5\left( w \cdot (u)/(3) \cdot (3)/(v)\right) \\= \log_5\left( (wu)/(v)\right)

User Pablo Lopez
by
7.7k points

Related questions

asked Jan 15, 2024 184k views
Soveran asked Jan 15, 2024
by Soveran
8.4k points
1 answer
0 votes
184k views
asked Jul 7, 2024 158k views
Brandon Wagner asked Jul 7, 2024
by Brandon Wagner
8.4k points
1 answer
2 votes
158k views
asked Jun 5, 2024 32.5k views
Msell asked Jun 5, 2024
by Msell
7.4k points
1 answer
2 votes
32.5k views