All About Pocket Knives

Jerry I do believe that's what they call "double trouble". Please correct me if I'm wrong.

Phil, to figure this one you have to use a logarithmic equation so it's actually more than "double trouble".

jerryd6818 wrote:Phil, to figure this one you have to use a logarithmic equation so it's actually more than "double trouble".

Here you go.
Solve 2logb(x) = logb(4) + logb(x – 1)
All of these logs have the same base, but I can't solve yet, because I don't yet have "log equals log". So first I'll have to apply log rules:

2logb(x) = logb(4) + logb(x – 1)

logb(x2) = logb((4)(x – 1))
logb(x2) = logb(4x – 4)

Then:

x2 = 4x – 4
x2 – 4x + 4 = 0
(x – 2)(x – 2) = 0

The solution is x = 2.

Smarty pants. I wouldn't know a mathematical log if it bit me in the butt.

..................that makes 2 of us, jerry!!!............... ..........................