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[carjon011208]

I have been trying to figure out this problem forever
loga(x-1)-loga(x+6)=loga(x-2)-loga(x+3)

I end up with [(x-1)/(x+6)] = [(x-2)/(x+3)] but i can't figure out where to go from there.

All i did was eliminate the logs, it's the first step, I just don't know where to go from there, it's as far as i can get.

 

[mmm4444bot]

Hi Carjon:

I'm not sure how you arrived at your equation because you did not show your work.

If you had written log[sub:25ed461y]a[/sub:25ed461y][(x - 1)/(x + 6)] on the left-hand side of your equation, then I would understand that side.

Use the following property on both sides of the original equation.

log(A) - log(B) = log(A/B)

Then you can write an equation by setting the arguments of the logarithms equal.

Solve that equation for x.

If you need more help, then please show your work. It also helps us to help you if you ask questions about why you're stuck.

Cheers,

~ Mark

 

[skeeter]

loga(x-1)-loga(x+6)=loga(x-2)-loga(x+3)

I end up with [(x-1)/(x+6)] = [(x-2)/(x+3)] but i can't figure out where to go from there.

All i did was eliminate the logs, it's the first step, I just don't know where to go from there, it's as far as i can get.

\(\displaystyle \frac{x-1}{x+6} = \frac{x-2}{x+3}\)

cross-multiply ...

\(\displaystyle (x-1)(x+3) = (x+6)(x-2)\)

expand both sides, collect like terms, and solve for x. remember to check your solutions in the original log equation.

 

[mmm4444bot]

... All i did was eliminate the logs, ...

No. You also did something else. You made a huge typographical error in your original post. Now that you've corrected this error, your equation makes sense.

Follow Skeeter's instructions.

If you need more assistance, then please show your work.

Cheers,

~ Mark