2007-08-22 18:53:01 · 4 answers · asked by Hanna 2 in Science & Mathematics ➔ Mathematics
can u show it by just using log (not using ln)?
2007-08-22 19:06:35 · update #1
8 (2^2x) = 3 + 2 (2^x)
Let y = 2^x
8y² - 2y - 3 = 0
(4y - 3) (2y + 1) = 0
y = 3 / 4 , y = - 1 / 2
2^x = 3 / 4 , 2^x = (- 1/2)
Taking + ve value as being valid:-
x log 2 = log(3 / 4)
x = log (3/4) / log 2
x = (- 0.125) / (0.3)
x = - 0 . 417
2007-08-22 19:59:02 · answer #1 · answered by Como 7 · 2⤊ 0⤋
2^(2x+3) = 3 + 2^(x+1)
(2x+3)ln[2]=ln[3]+(x+1)ln[2]
(2x+3)ln[2]-(x+1)ln[2]=ln[3]
ln[2]((2x+3)-(x+1))=ln[3]
2x+3-x-1=ln[3]/ln[2]
x+2=ln[3]/ln[2]
x=ln[3]/ln[2]-2
x is approx. = -0.415037
2007-08-22 19:04:13 · answer #2 · answered by Anonymous · 0⤊ 2⤋
zero
2007-08-22 19:03:12 · answer #3 · answered by Anonymous · 0⤊ 3⤋
not equal.....
2007-08-22 19:06:27 · answer #4 · answered by John C 1 · 0⤊ 2⤋