My Math Forum solving for x in logarithmic equations.

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 December 6th, 2008, 07:29 PM #1 Newbie   Joined: Dec 2008 Posts: 1 Thanks: 0 solving for x in logarithmic equations. Hello, I have these set of questions given, I would like to know how to solve the equations (for x), and the process. for instnace, 2logx=3log4 , where as the base of both logarithims is generally 10 in this case, I know x=8 but how do we find this, then it gets tricker for questions like this.. logx+log3= log12 log (base 2) (x+2) + log(base 2) x = 3 log (base 2) (x-2) + log (base 2 ) (x+1) = 2 If log (base 2) (log base 2) (a) = 2, what is a For the first 2, do we change them into exponential form first, I can work my way through a) and b) but im not sure for the rest, and I don't seem to find a quick efficient way for the first 2. Any Rules I could apply to, 2logx=3log4, that work universally for the other 4 questions too?
 December 7th, 2008, 03:48 AM #2 Global Moderator   Joined: Dec 2006 Posts: 20,390 Thanks: 2015 I'll assume you're interested only in real solutions. I'll use "iff" to mean "if and only if" and "lg" to mean "log (base 2)". 2log(x) = 3log(4) iff log(x^2) = log(4^3) and x > 0, so x = (4^3)^(1/2) = 8 is the only real solution. log(x) + log(3) = log(12) iff log(3x) = log(12), so x = 12/3 = 4 is the only real solution. lg(x+2) + lg(x) = 3 = lg( iff lg(x(x + 2)) = lg( and x > 0, i.e., x^2 + 2x + 1 = 9 and x > 0, so x = 9^(1/2) - 1 = 2 is the only real solution. lg(x - 2) + lg(x+1) = 2 = lg(4) iff lg((x - 2)(x + 1)) = lg(4) and x > 2, i.e., x^2 - x + 1/4 = 6 1/4 and x > 2, so x = 3 is the only real solution. lg(lg(a)) = 2 = lg(4) iff lg(a) = 4 iff a = 2^4 = 16.
 December 7th, 2008, 12:48 PM #3 Newbie   Joined: Nov 2008 Posts: 10 Thanks: 0 Re: solving for x in logarithmic equations. Use the laws of logarithms. For example: Should be very usable.
 December 7th, 2008, 04:21 PM #4 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,923 Thanks: 1123 Math Focus: Elementary mathematics and beyond Re: solving for x in logarithmic equations. ooops...

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