1. ## Log Equation

Okay, this is a (for you guys)simple Log Equation, the final is in a few day's and i've forgotten how to solve this. Forgotten some of the rules.

I'm stuck here bty. - logx = log5 + log3 + log2

2. Originally Posted by butters
Okay, this is a (for you guys)simple Log Equation, the final is in a few day's and i've forgotten how to solve this. Forgotten some of the rules.

I'm stuck here bty. - logx = log5 + log3 + log2
$\displaystyle \log{x} - \log(x-2) = \log{5} - \log(x+3)$

$\displaystyle \log\left(\frac{x}{x-2}\right) = \log\left(\frac{5}{x-3}\right)$

$\displaystyle \frac{x}{x-2} = \frac{5}{x-3}$

finish up and solve for x ... don't forget to check your solution(s) in the original equation for domain issues.

3. Hi there butters, consider the log law

$\displaystyle \ln(a)-\ln(b)= \ln\left(\frac{a}{b}\right)$

so

$\displaystyle \ln(x)-\ln(x-2)= \ln(5)-\ln(x+3)$

$\displaystyle \ln\left(\frac{x}{x-2}\right)= \ln\left(\frac{5}{x+3}\right)$

$\displaystyle \frac{x}{x-2}= \frac{5}{x+3}$

can you take it from here?

4. Originally Posted by skeeter
finish up and solve for x ... don't forget to check your solution(s) in the original equation for domain issues.
I take it the initial steps i was taking in "logx = log5 + log3 + log2" were wrong..
But i do remember solving problems like this in that method.
Thank You.
Originally Posted by pickslides

can you take it from here?
Yes, thank you, i'm getting "1 + or - 3i".
The answer is given as "No solution". So does that mean, like negative number answers, the "i" make's it a "No solution" ?

5. It means no real solution.

6. Okay, once again, Thanks.