Please help me figure out this problem...how do I set it up?
2007-10-08 06:50:26 · 5 answers · asked by flatfoot 3 in Science & Mathematics ➔ Mathematics
first: ln(2x-3)=-5/3
then 2x-3=exp(-5/3)
<=> x=[exp(-5/3)+3]/2
2007-10-08 06:55:29 · answer #1 · answered by Anonymous · 2⤊ 2⤋
easy
3*ln(2x-3)= -5
First divide by 3
ln(2x-3)= -5/3
e both sides to eliminate ln
e *ln (2x-3) = e^(5/3)
e and ln cancel out
2x-3= e^(-5/3)
add 3 both sides
2x= 3+e^(-5/3)
divide by 2
x=
3+ e^(-5/3)
------------
2
plug it in in the calculator you should get 2.93
2007-10-08 13:53:04 · answer #2 · answered by Anonymous · 1⤊ 3⤋
3*LN(2x-3)= -5
=> LN(2x-3)= -5/3
=> 2x-3= e^(-5/3)
=> 2x= e^(-5/3)+3
=> x= [ e^(-5/3)+3 ] / 2
2007-10-08 13:55:49 · answer #3 · answered by harry m 6 · 0⤊ 4⤋
ln (2x - 3) = - 5/3
2x - 3 = e ^(- 5/3)
2x = 3 + e^(- 5/3)
x = 3/2 + (1/2)(1/ e^(5/3))
x = 1.5 + (0.5)(0.189)
x = 1.59 (to 2 decimal places)
2007-10-08 13:57:05 · answer #4 · answered by Como 7 · 1⤊ 4⤋
natural logs have the base "e," and the coefficient is the power of the rest, so:
e^(-5)=(2x-3)^3
x=[cuberoot(e^(-5))-3]/2
2007-10-08 13:54:47 · answer #5 · answered by Amelia 6 · 0⤊ 4⤋