Solve the following logarithmic equations:
a) log2 (6x-8) - log2 (2x+4) = log2 (x-6)
b) log3 (x+3) - log3 (x-3) = 1
c) logn 1 = x
kinda stuck on those ones
............since
Now we can just drop the logs:
I think you can take it from here
b) log3 (x+3) - log3 (x-3) = 1
...........since if , then
...and i think you can solve that equation
right off the bat!c) logn 1 = x
The always, i don't care what the base is, the log of 1 is 0, why?
Anything raised to the zero gives one, so the power is zero
Hello, kevinz!
Solve the following logarithmic equation:
Here's a trick I've taught all my students . . .
If we have a logarithmic equation and a log-term is negative,
. . "move" it to the other side. .**
So we have: .
. . . . .Then: .
Exponentiate (or "un-log") both sides: .
We have a quadratic: . .
. . . . . which factors: .
. . . . .and has roots: . . . . . -1 is extraneous
Therefore, the only root is: .
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
** .Why? .To avoid that annoying quotient.