# Thread: Solving a Log Equation

1. ## Solving a Log Equation

log6(x+1) + log6(x+2) = 1

The 6 in both equations is the base.

I know that...

log6(x+1) + log6(x+2) = log6(x+1)(x+2)

So I get to here.

log6(x+1)(x+2) = 1

I assume we have to make it equal to 0.

log6(x^2+3x+2) = 1
log6(x^2+3x+1) = 0

Tell me if I'm doing something wrong here.

The answer is supposed to be 1, but I don't understand how to get to the answer.

2. Originally Posted by Jeavus
log6(x+1) + log6(x+2) = 1

The 6 in both equations is the base.

I know that...

log6(x+1) + log6(x+2) = log6(x+1)(x+2)

So I get to here.

log6(x+1)(x+2) = 1

I assume we have to make it equal to 0.

log6(x^2+3x+2) = 1
log6(x^2+3x+1) = 0
that's wrong:

remember the definition of a logarithm... If $\displaystyle \log_a b = c$ then $\displaystyle a^c = b$

so you have: $\displaystyle \log_6 (x^2 + 3x + 2) = 1$

$\displaystyle \Rightarrow 6^1 = x^2 + 3x + 2$

now continue

3. Thanks again.

4. How do I go about solving this equation?

log5(2x+2) - log5(x-1) = log5(x+1)

The 5 in each is the base.

5. Use the formula $\displaystyle \displaystyle\log_ab-\log_ac=\log_a\frac{b}{c}$.

6. Hmmm...

log5(2x+2) - log5(x-1) = log5(x+1)
log5(2x+2/x-1) = log5(x+1)
log5(2x+2/x-1)(x+1) = 0

Where do I go from here?

7. Originally Posted by Jeavus
Hmmm...

log5(2x+2) - log5(x-1) = log5(x+1)
log5(2x+2/x-1) = log5(x+1)
log5(2x+2/x-1)(x+1) = 0

Where do I go from here?
how did you get $\displaystyle \log_5 \left( \frac {2x + 2}{x - 1} \right)(x + 1)$ exactly? you had to subtract $\displaystyle \log_5 (x + 1)$ correct?

when you find what the log should be, follow what i told you in the first post

...or, you could have simply dropped all the logs...

$\displaystyle \log_5 \left( \frac {2x + 2}{x - 1}\right) = \log_5 (x + 1)$

$\displaystyle \Rightarrow \frac {2x + 2}{x - 1} = x + 1$ ............do you see why?

8. $\displaystyle \displaystyle\log_5\frac{2x+2}{x-1}=\log_5(x+1)\Rightarrow\frac{2x+2}{x-1}=x-1$
Now, solve for x.

9. Thanks very much.