How do you solve 6e^(-4k-10)-4=63?

2 Answers
Jul 25, 2018

I tried this:

Explanation:

Let us rearrange it a bit:

e^(-4k-10)=(63+4)/6

now take the natural log of both sides:

ln[e^(-4k-10)]=ln[(63+4)/6]

we can get rid of ln and e on the left and write:

-4k-10=ln[(63+4)/6]

rearrange to isolate k:

4k=-10-ln[(63+4)/6]

and:

k=1/4[-10-ln((63+4)/6)]=-3.10323

k=-1/4(10+\ln(67/6))=-3.1032

Explanation:

Given that

6e^{-4k-10}-4=63

6e^{-4k-10}=67

e^{-4k-10}=67/6

taking \ln on both the sides as follows

\lne^{-4k-10}=\ln(67/6)

(-4k-10)\ln e=\ln(67/6)

-4k-10=\ln(67/6)\quad (\because \ln e=1)

4k=-10-\ln(67/6)

k=-1/4(10+\ln(67/6))

=-3.1032