Logarithmic problem help

Apr 2010
21
0
I have to solve x in terms of k

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

HallsofIvy

MHF Helper
Apr 2005
20,249
7,909
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?
 
Apr 2010
21
0
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.
 
Nov 2009
927
260
Wellington
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 \(\displaystyle k\).