How would i go about solving for x in this equation 3e^(-2x) =5 ?

Question

e^ is the way i wrote e raised to (-2x)

Answer 1

the equation is 3e^(-2x) = 5
Taking log on both sides we get
log(3e^-2x) = log5
log3 + log(e^-2x) = log5
log3 - 2xloge = log5
log3 - 2x = log5 (since loge = 1)
or 2x = log3 - log5
= log3/5
x = 1/2(log3/5)

Answer 2

3e^(-2x) = 5

divide both sides by 3: e^(-2x) = 5/3

take the Natural Log of both sides: LN [e^(-2x) ] = LN [ 5/3]

by the Inverse Function Theorem: -2x = LN [ 5/3 ]

Divide both sides by -2: x = (LN [ 5/3]) / (-2)

you can get this final answer from a scientific calculator

remember that these types of problems can potentially produce extraneous (ie false) solutions. Always check solutions by plugging back in to the ORIGINAL equation to make sure it works. Be especially careful with Log equations since the domain of Log functions are (0, +Infinity). You cant plug a negative x value into a log function.

Answer 3

First, divide by 3:
e^(-2x) = 5/3
Then take the natural log of both sides:
ln(e^(-2x)) = ln (5/3) = -2x
then divide by -2
x = -ln(5/3)/2

Answer 4

3e^(-2x)=5
ln 3e^(-2x) = ln 5
ln 3 + ln e^(-2x) = ln 5
ln 3 - 2x = ln 5
2x = ln 3 - ln 5
2x = ln (3/5) = -0.51
x = -0.51 / 2 = -0.255

Answer 5

e^(-2x)=5/3
take natural log on both sides,

-2x=ln(5/3)

-2x=0.511

x= -0.511/2

x=-0.2555

Answer 6

3*(-2X) ln e= ln 5 (ln=natural log); where: ln e = 1

-6x=1.61

x=-0.27

Answer 7

i would solve it for you but i have never heard of "e^" in a equation. sorry.