Trouble with the number e?

Question

solve for x:

2e^(2x) -e^(-x) = 0

show how you used the natural log to solve it.

Answer 1

log_b(mn) = log_b(m) + log_b(n)
log_b(m/n) = log_b(m) – log_b(n)
log_b(mn) = n · log_b(m)
ln (e^x) = x
ln (1) = 0
ln (0) = cannot be determined


2e^(2x) -e^(-x) = 0
2e^(2x) = e^(-x)

multiply both sides with ln
ln [2e^(2x)] = ln [e^(-x)]
ln(2) + ln [e^(2x)] = -x
ln(2) + 2x = -x

add (-2x) on both sides
ln(2) + 2x -2x = -x -2x
ln(2) = -3x

divide both sides with -3
ln(2) / (-3) = -3x / (-3)
x = - ln(2) / 3

Answer 2

Multiply by e^x:
2e^(3x) = 1
e ^(3x) = 1/2
Now take the natural log
of both sides:
3x = log(1/2) = log 1 - log 2 = -log 2
x = -log 2/3.

Answer 3

x = 0
you take the ( Ln) for the equation it gose like
2x ln 2e - x ln e = 0
ok now ln e = 1
2x (ln 2 X 1 ) - x(1) = 0
=> 2x X 0.693 - x = 0
so 0.386x =0
then x = 0

Answer 4

the trick with e is that ln(e) = 1 so

start with 2e^(2x) = e^(-x)

then take log of both sides

ln( 2e^(2x) ) = ln( e^(-x) )

ln(2) + ln( e^(2x) ) = ln (e^ (-x)

ln2 + 2x = -x

ln(2) = -3x

then

x = -ln(2) / 3

Answer 5

Yes some how the I's end up looking like an h or something and when you say decipher well you can never understand it and in your situation well that would be worse you have no option that is horrible! Those things are horrible

Answer 6

ln(2e^(2x)) - ln(e^(-x)) = ln(0)

2xln(2e) + xln(e) = 0

2xln(2) + 1 + X = 0

x(2ln(2) +1) = -1

x = -1/(2ln(2) +1)

Answer 7

its not a number its like a constant, it cancels out with ln. just use basic algebra and raise by ln or e when u have to, then solve for x. when u solve for x, it has to equal 0,remember that, k

natural log is ln _____/ln _______ or log______/log______ thats all

Answer 8

2e^(2x) = e^-x
ln(2e^(2x)) = ln(e^-x)
ln 2 + 2x = -x
ln 2 = -3x
x = ln2 /-3

Answer 9

2e^(2x) - e^(-x) = 0
2e^(3x) - 1 = 0
e^(3x) = 1/2
3x = -ln2
x = -1/3 ln2

Answer 10

2e^(2x) - e^(-x) = 0
2e^(3x) - 1 = 0
e^(3x) = 1 / 2
e^(3x) = 0.5
3x ln e = ln (0.5)
3x = ln (0.5)
x = ln (0.5) / 3