Asked by Abraham
solve the equations for x
a) e^2x - e^x - 30 = 0
b)1-3/n^x = 5
a) e^2x - e^x - 30 = 0
b)1-3/n^x = 5
Answers
Answered by
Damon
let z = e^x
then
z^2 - z - 30 = 0
(z-6)(z+5) = 0
so
z = 6 or -5
if z = e^x = 6
x = ln 6 look it up
if z = -5
x = ln -5 impossible so only ln 6 is solution
then
z^2 - z - 30 = 0
(z-6)(z+5) = 0
so
z = 6 or -5
if z = e^x = 6
x = ln 6 look it up
if z = -5
x = ln -5 impossible so only ln 6 is solution
Answered by
Damon
Are you sure this is not a typo?
1-3/n^x = 5
-3/n^x = 4
n^x = -3/4
ln n^x = -ln .75
x ln n = -ln.75
x = -ln.75/ln n
x = .2877/ln n
1-3/n^x = 5
-3/n^x = 4
n^x = -3/4
ln n^x = -ln .75
x ln n = -ln.75
x = -ln.75/ln n
x = .2877/ln n
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.