(log2X)^2 - log2X^2 + 2 =0
2007-11-25 05:58:35 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
and also this one:
3(log3X)^2 - log3X^28 + 9=0
2007-11-25 06:00:41 · update #1
(log2X)^2 - log2X^2 + 2 =0
Let u = log 2x
Then u^2 -2u +2 = 0
u =[2 +/- sqrt(2^2-8)]/2
u = 1 +/-2i
log 2x = 1 +/- 2i
x = .5*10^(1 +/- 2i)
The other problem is done similarly.
2007-11-25 06:10:21 · answer #1 · answered by ironduke8159 7 · 0⤊ 0⤋
Use the rule log(a*b) = log(a) + log(b). You can then simplify your equation to:
(log x + log 2)^2 - [ log 2 + 2 logx] + 2 = 0
(log x)^2 + (2 log 2) log x + (log 2)^2 - log 2 - 2 log x + 2 = 0
Re-arranging, you get:
(log x)^2 + (2 log 2 - 2) log x + [(log 2)^2 - log 2 + 2] = 0
From here, substitute y = log x, then use the quadratic formula to solve for y, then get x from x = exp(y).
2007-11-25 14:12:19 · answer #2 · answered by Dave A 3 · 0⤊ 0⤋
holy ****, you dont
2007-11-25 14:06:28 · answer #3 · answered by jenniemae 2 · 0⤊ 0⤋