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

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Answer 1 · 2018-07-25T13:32:24

I tried this:

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#

Answer 2 · 2018-07-25T13:32:43

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

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#