How do I solve this algebraically? -5e^-x + 9 = 6?

Answer 1

-5e^-x + 9 = 6 ==> minus 9 each side
-5e^-x = -3 ==> divide by -5
e^-x = 3/5 ==> ln each side {recall ln(e^x) = x }
-x = ln(3/5)
x = - ln(3/5) ==> which is equal to ln(5/3)


=]

Answer 2

step by step solution:

-5e^-x + 9 = 6
-5e^-x = 6 - 9
e^-x = (6 - 9)/-5
- x root of e = ((6 - 9)/-5)^-x

The last part, there can be a few answers because there is no definate value of -x and e

if you knew the value of one of them, then you can continue to solve the equasion

Answer 3

-5e^-x + 9 = 6
<=> -5 e^(-x) = -3
<=> e^(-x) = 3/5
<=> -x = ln (3/5)
<=> -x = ln 3 - ln 5
<=> x = ln 5 - ln 3.

Answer 4

take log and it will become very easy and then take antilog