300e^-.076t = 100

[wilc0919]

I put e^-.076t = 1/3

-.076t = ln 1/3

t= ln 1/3 OVER -.076 EQUALS 14.455

Is that right????

 

[stapel]

Is your exercise in the subject line? (Subject lines frequently do not relate to posts, so please forgive me if this assumption is in error.) If so, what are the instructions? Also, is "t" inside the exponent, or not?

If the exercise is a "solving" one, then check your answer by plugging it back into the original exercise. If it works, then it's probably correct.

Eliz.

 

[wilc0919]

The problem is relating to natural logs

the problem is 300e^-.076t = 100 (yes, t is in the exponet)

so I put

e-.076t = 1/3

so -.076t = ln 1/3

so t = ln 1/3 over -.076 = 14.455

Can you just tell me if I am doing it right please....plugging back in is even harder at this point and I just want to know if I am even doing the problem right....thanks so much

 

[stapel]

the problem is 300e^-.076t = 100 (yes, t is in the exponet)

I will guess that the instructions are to "solve". (It is usually best to put the entire question within your post. This would include the instructions.)

so I put e-.076t = 1/3

I will guess that you mean "e^(0.076t) = 1/3", from having divided through by 300.

so -.076t = ln 1/3

so t = ln 1/3 over -.076 = 14.455

To check, plug the solution back in:

. . . . .300 e^-0.076t with t = 14.455 (approx.)

. . . . .300 e^{-0.076(14.455)}

. . . . .300 e^-10.09858

. . . . .(300)(0.3333440964...)

. . . . .100.0032289...

Allowing for round-off error, this is close enough to confirm the solution.

plugging back in is even harder at this point....

So you're working from tables? You're not allowed to use calculators? That's a bummer!

Eliz.