Hard problems for Advanced Algebra
2007-01-03 02:03:42 · 10 answers · asked by td 2 in Science & Mathematics ➔ Mathematics
log2 + 2logx = log(4x-2)
log2 + log(x^2) = log(4x-2) ....... since a*log b = log(b^a)
log[2*(x^2)] = log(4x-2) ........ since log a + log b = log a*b
2*(x^2) = 4x - 2
x^2 = 2*x - 1
x^2 - 2*x +1 = 0
(x - 1)^2 = 0
Two equal roots: x=1,1
Thus, x = 1
2007-01-03 02:06:51 · answer #1 · answered by Som™ 6 · 1⤊ 0⤋
log2 + 2logx = log(4x-2) log2 + logx^2 = log(4x-2) log(2x^2) = log(4x-2) 2x^2 = 4x-2 2x^2 - 4x + 2 = 0 2(x^2 - 2x + 1) = 0 2(x-1)(x-1) = 0 x = 1
2016-05-22 22:43:32 · answer #2 · answered by Anonymous · 0⤊ 0⤋
log2 + 2logx = log(4x-2)
log2 + log(x^2) = log(4x-2)
log[2*(x^2)] = log(4x-2)
2*(x^2) = 4x - 2
Dividing both sides by 2
x^2 = 2*x - 1
x^2 -2x +1 = 0
x^2 -2.x.1 +1^2 = 0
(x - 1)^2 = 0
(x - 1)(x - 1) = 0
(x - 1) = 0, (x - 1) = 0
x = 1, 1
2007-01-03 02:13:44 · answer #3 · answered by Sheen 4 · 0⤊ 0⤋
log2+2logx=log(4x-2)
log2+log(x^2)=log(4x-2)
{2x^2/(4x-2)}=0
2x^2=4x-2
2x^2-4x+2=0
x^2-2x+1=0
(x-1)^2=0
So, x=1
2007-01-03 03:25:25 · answer #4 · answered by sugan 1 · 0⤊ 0⤋
here is the solution:
=>Log2 + 2log x=log(4x-2)
=>Log2 +logx2=log(4x-2) //by the property of log
=>2 * x2 =4x – 2
=>2 x2 – 4x +2=0
=>x2 – 2x +1 =0
=>(x-1)2=0
=>x = 1,1
2007-01-03 02:23:26 · answer #5 · answered by Anonymous · 0⤊ 0⤋
log2 + 2logx = log(4x-2)
log 2x^2=log (4x-2)
2x^2=4x-2
x^2-2x+1=0
(x-1)^2
2007-01-03 03:45:10 · answer #6 · answered by yupchagee 7 · 0⤊ 0⤋
2 log x = log x²
log 2 + log x² = log 2x²
So log (2x²) = log (4x-2)
2x² = 4x-2
=> 2x² - 4x + 2 = 0
=> x² - 2x + 1 = 0
so x = 1 (1 is a double root)
2007-01-03 02:11:47 · answer #7 · answered by anton3s 3 · 0⤊ 0⤋
x=56
2007-01-03 02:10:30 · answer #8 · answered by Anonymous · 0⤊ 0⤋
log2x^2=log(4x-2)
2x^2=4x-2
2x^2-4x+2=0
dividing by2
x^2-2x+1=0
(x-1)^2=0
x=1,1
2007-01-03 02:09:37 · answer #9 · answered by raj 7 · 0⤊ 0⤋
Hey I came up with x=2.... lemme know if I am right... i dont even know for real if I am ..... but hopefully I am!
2007-01-03 02:09:38 · answer #10 · answered by Get_in_my_belly 3 · 0⤊ 0⤋