# Logarithmic problem help

• Jul 13th 2010, 10:50 AM
Nikhiln25
Logarithmic problem help
I have to solve x in terms of k

\$\displaystyle log x + log (x+9)=k\$
• Jul 13th 2010, 10:55 AM
Also sprach Zarathustra
log(x) + log(y) = log(xy)
• Jul 13th 2010, 10:57 AM
Nikhiln25
Quote:

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
• Jul 13th 2010, 12:36 PM
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?
• Jul 13th 2010, 04:25 PM
Nikhiln25
Quote:

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.
• Jul 13th 2010, 11:04 PM
Bacterius
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\$.