Help with: log (x+90) - log (x+2) = log x

[otterpop]

Can anyone point me to a tutorial or explain how to solve an equation like this? I have found some helpful information on solving logarithmic equations but nothing that has "log x" at the end. This website has been good - http://www.purplemath.com/modules/solvelog2.htm - but I still can't figure out this one with the info they provide. Most of the examples that I've found have a whole number at the end.

log (x+90) - log (x+2) = log x
Answer: 9

Thanks so much

 

[otterpop]

I think this is easier than you think

\(\displaystyle \begin{align*}&\log(x+90)-\log(x+2)=\log x \\ &log(\frac{x+90}{x+2})=\log x \\ &\frac{x+90}{x+2}=x \\ &x+90=x^2+2x \\ &0=x^2+x-90 \\ &(x+10)(x-9)=0 \\ &x=-10\;\; x=9\end{align*}\)

but for real numbers \(\displaystyle \log(-10+2)=\log(-8)\) is not defined so x=9 only.

Thank you so much! That makes sense.

 

[srmichael]

Thank you so much! That makes sense.

Romsek used one of the Big 3 Log Rules. Know these three little diddys and you will survive many a log problems.

\(\displaystyle log_{b}(u \cdot v) = log_{b}u+log_{b}v\)

\(\displaystyle log_{b}(\dfrac{u}{v}) = log_{b}u-log_{b}v\)

\(\displaystyle log_{b}(u^v) = vlog_{b}u\)

 

[HallsofIvy]

You also need the fact that the logarithm is a "one to one" function: if log(x)= log(y) then x= y.