solve the equation 6^(4x-3) = 36^(x+1)
2007-12-09 17:54:29 · 6 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
6^(4x-3) = 36^(x+1)
6^(4x-3) = (6^2)^(x+1)
6^(4x-3) = 6^2(x+1)
(4x-3) = 2(x+1)
4x-3 = 2x+2
2x = 5
x = 5/2
2007-12-09 17:57:27 · answer #1 · answered by I.prefer.gmail 4 · 0⤊ 1⤋
36 = 6^2
6^(4x - 3) = (6 ^ 2)^(x + 1)
6^(4x - 3) = 6^[2(x + 1)]
4x - 3 = 2(x + 1)
4x - 3 = 2x + 2 // -2x
2x - 3 = 2 // +3
2x = 5 // divide by 2
x = 2.5
2007-12-09 17:59:26 · answer #2 · answered by Amit Y 5 · 0⤊ 0⤋
6^(4x - 3) = 6^( 2(x + 1) )
Take log base 6:-
(4x - 3) log 6 = 2(x + 1) log 6
4x - 3 = 2x + 2
2x = 5
x = 5 / 2
2007-12-09 20:30:06 · answer #3 · answered by Como 7 · 2⤊ 0⤋
you write 36^(x+1)=6^[2(x+1)]
so
6^(4x-3)= 6^(2x+2)
4x-3 =2x+2
x=2.5
2007-12-09 18:01:26 · answer #4 · answered by Oana 3 · 0⤊ 0⤋
section a million: L=(36x^4-64x^2)/(6c^2+8x) section 2: that's a distinction of squares. There are basically 2 words and that they are the two squares and the sign interior the middle is unfavourable. section 3: That divides relatively particularly. it would be L=6x^2-8x. section 4 is the comparable element, interestingly like.
2016-10-10 23:17:26 · answer #5 · answered by loy 4 · 0⤊ 0⤋
use Log rule or the 36^x=6^2x rule.
2007-12-09 17:57:05 · answer #6 · answered by Nostradamus is back squared 3 · 0⤊ 0⤋