salvage value after t yars given by V = 35000 e^(-0.43t)

[carson_6969]

I've been trying to figure this problem help Here is the question

A business estimates that the salvage value V of a piece of machinery after t years is given by.

V(t)= $35000e^(-0.43t)

What amount of time will the salvage value be $7771?

I know i have to find t but i have been trying and here is the work that i came up with

7771= 35000e^(-0,43t)

*-27229=e^(-0,43t)

ln -27229= ln e^(-0,43t)

ln -27229 = (-0,43t)

ln -27229 / (-0,43) = t

23 = years

help please

 

[galactus]

You subtracted 35000 from 7771. No. Don't you have to divide?.

7771/35000

 

[Deleted member 4993]

Re: I've been trying to figure this problem help

Here is the question

A business estimates that the salvage value V of a piece of machinery after t years is given by.

V(t)= $35000e^(-0.43t)

What amount of time will the salvage value be $7771?

I know i have to find t but i have been trying and here is the work that i came up with

7771= 35000e^(-0,43t)
]
7771/35000 = e^(-0.43t)

35000/7771 = e^(+0.43t)

4.503924849 = e^(+0.43t)

.43 * t = ln(4.503924849) = 1.504949205


t = 1.504949205/.43 = 3.499881873 = 3.5


*-27229=e^(-0,43t)

ln -27229= ln e^(-0,43t)

ln -27229 = (-0,43t)

ln -27229 / (-0,43) = t

23 = years

help please