I have to solve x in terms of k

$\displaystyle log x + log (x+9)=k$

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- Jul 13th 2010, 10:50 AMNikhiln25Logarithmic problem help
I have to solve x in terms of k

$\displaystyle log x + log (x+9)=k$ - Jul 13th 2010, 10:55 AMAlso sprach Zarathustra
log(x) + log(y) = log(xy)

- Jul 13th 2010, 10:57 AMNikhiln25
- Jul 13th 2010, 12:36 PMHallsofIvy
Yes, Also Sprach Zarathustra understood that. Apply his hint. If "log(x)+ log(y)= log(xy)" is a general rule, then what is log(x)+ log(x+ 9)?

Do you know how to solve log(y)= k for y? Do you know what the**inverse**function to log(x) is? - Jul 13th 2010, 04:25 PMNikhiln25
I haven't taken pre calc in 3 years so my memory of it is really sketchy..this is a review problem for calc which I'm taking right now.

Log(x^2+9x)=k

10^k=(x^2+9x)

Can't really figure out where to go from there. Like I said I barely remember anything from pre calc since its been about 3 years. - Jul 13th 2010, 11:04 PMBacterius
That's the right way to do it, now notice that it is equal to the equation $\displaystyle x^2 + 9x - 10^k = 0$, which is a quadratic equation in $\displaystyle x$ with a constant term in $\displaystyle k$. Therefore you can apply the quadratic formula to isolate $\displaystyle x$ and solve it in terms of $\displaystyle k$.