Question

algebra 2 help

Answer 1

which one you want to do first? the second is easiest

Answer 2

the second one

Answer 3

\[e^{-5x}+10=16\\
e^{-5x}=6\] is a start
then write in equivalent logarithmic form as
\[-5x=\ln(6)\] and finally
\[x=-\frac{\ln(6)}{5}\]

Answer 4

really thats all for the second one? how do i do the first one?

Answer 5

yes, that is all
let me open it again and look

Answer 6

\[\large 4^{2t}=5^{4t+4}\] right?

Answer 7

this is going to require one log step, then a raft of algebra
much harder than the second one
ready?

Answer 8

log step is to rewrite it as
\[2t\ln(4)=(4t+4)\ln(5)\] then the algebra steps are to solve for \(t\)

Answer 9

realizing of course that \(\ln(4)\) and \(\ln(5)\) are just numbers (constants) you have
\[2t\ln(4)=4t\ln(5)+4\ln(5)\] or
\[2t\ln(4)-4t\ln(5)=4\ln(5)\] factor out the \(t\) and get
\[t(2\ln(4)-4\ln(5))=4\ln(5)\] and finally divide to get
\[t=\frac{4\ln(5)}{2\ln(4)-4\ln(5)}\] i guess you can cancel a 2 top and bottom and finish with
\[t=\frac{2\ln(5)}{\ln(4)-2\ln(5)}\]

Answer 10

oh ok thank you so much!

Answer 11

yw