Question 24375
Hi,

It sounds like you need a lesson on logorithms, rather than help with a specific problem. Try <a href="http://www.bbc.co.uk/education/asguru/maths/13pure/05exposandlogs/index.shtml">these</a> web pages. I've heard that BBC bitesize is really good. See how you get on with that, and then try to follow my solution of your problem. If your still having trouble then please write back and I'll see what I can do.

We want to solve

*[tex \log_2 x(x-4) = 5]

Now when solving an equation for x, we try to undo everything that's been done to x, but we undo it in reverse. For example To solve *[tex 2x+3=9] we first undo the plus 3, by subtracting 3, then we undo the times 2 by dividing by 2.

To undo logorithm we raise the base to the power of each side. In our case the base is two. So

*[tex 2^{\log_2 x(x-4)}=2^5]

Now remeber that this undoes the logorithm, so the left gand side of the equation becomes *[tex x(x-4)]. The right hand side is obviously 32. So we are now left with the quadratic

*[tex x^2-4x=32]

You can solve this by any method but I prefer completing the square. Regardless of how you do it, the answers are -4 and 8.

The final thing you need to remember is that logs only work for posotive numbers, so your solutions are only valid if they satisfy x(x-4)>0. Check this!

Hope that helped, if not please do ask again, and tell me the first bit about logorithms that confuses you, and we can work on it.

Kev