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

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- Jun 22nd 2007, 08:47 PMkevinzLogarithm questions
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 - Jun 22nd 2007, 09:03 PMJhevon

............since

Now we can just drop the logs:

I think you can take it from here

Quote:

b) log3 (x+3) - log3 (x-3) = 1

...........since if , then

...and i think you can solve that equation

Quote:

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 - Jun 23rd 2007, 09:15 PMSoroban
Hello, kevinz!

Quote:

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.

- Jun 23rd 2007, 09:28 PMJhevon