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Topic review (newest first)

kylekatarn
2005-09-13 11:23:35

hi!
Its very simple indeed... I bet you stopped trying when you saw the x....if you continue with the simplifications you end up with a easy quadratic.

Since we have all logs @ same base we don't need the info "using a base of 3"

And because I'm a bit lazy today (:|), let l(u) denote the logarithm of the expression u

2l(x+2) + l(x-1) = 3 l(x)
//I assume you know how to manipulate logs.
l[ (x+2) ] + l(x-1) = l(x)
l[ (x+2).(x-1) ] = l(x)

//if we have l(z)=l(w) we can solve [z=w] with [z and w >0]
(x+2).(x-1) = x; x>1

//now this is the fun part
(x+2).(x-1) = x
(x+4x+4).(x-1) = x
x+4x+4x-x-4x-4 = x
x+4x-x-4 = x
+x+3x-4 = +x
3x-4 = 0

//I told you it was easy!
3x-4 = 0
3x = 4
x = 4/3
x = (2√3)/3

//remember the original equation's condition?? X>1
//we discard the negative solution, so the answer to the problem is

x=(2√3)/3



//omg...solving problems like these one makes me remember how I deeply miss my good ol' math classes.;|

mikau
2005-09-13 08:58:32

2 log (x+2) + log (x - 1) = 3 log x            using a base of 3.

I'm supposed to solve for X. Usually I'm pretty good with these but this I just can't solve. Usually I have to factor, but I keep ending up with a 3rd degree polynomial which my mathbook has not taught me how to solve. So I'm expected to solve it another way. Help! :-(

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