1. ## Logarithmic problem help

I have to solve x in terms of k

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

2. log(x) + log(y) = log(xy)

3. Originally Posted by Also sprach Zarathustra
log(x) + log(y) = log(xy)

Sorry wrote that completely wrong, meant to say solve for X in terms of K

4. 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?

5. Originally Posted by HallsofIvy
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?
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.

6. 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 $\displaystyle k$.