Solving a Log Equation

**Jeavus** · September 30th 2007, 08:25 AM

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.

**Jhevon** · September 30th 2007, 08:31 AM

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 $\log_a b = c$ then $a^c = b$

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

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

now continue

**Jeavus** · September 30th 2007, 08:38 AM

Thanks again.

**Jeavus** · September 30th 2007, 08:56 AM

How do I go about solving this equation?

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

The 5 in each is the base.

**red_dog** · September 30th 2007, 09:00 AM

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

**Jeavus** · September 30th 2007, 09:16 AM

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?

**Jhevon** · September 30th 2007, 09:18 AM

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 $\log_5 \left( \frac {2x + 2}{x - 1} \right)(x + 1)$ exactly? you had to subtract $\log_5 (x + 1)$ 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...

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

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

**red_dog** · September 30th 2007, 09:19 AM

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

**Jeavus** · September 30th 2007, 09:22 AM

Thanks very much.

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