Find unkown values b and B when y=e^x+b + B where points (0,2) & (1,6) are given.?

Question

Please note: ^ means to the power of.

This is a logarithm so e is Euler's number (2.718...)

It would really help if you could show the working out because I know the answer.

The two unknown values should be 0.84 and -0.33.

Answer 1

y = e^(x + b) + B

To solve this, form two equations and two unknowns by plugging each of the given points.

For (0, 2): x = 0, y = 2, so
2 = e^(0 + b) + B
2 = e^b + B; therefore,
B = 2 - e^b

For (1, 6): x = 1, y = 6, so
6 = e^(1 + b) + B, so
B = 6 - e^(1 + b)

Now that we've expressed B in two different ways, it must follow they are equal to each other. Equate them.

2 - e^b = 6 - e^(1 + b)
e^(1 + b) - e^b = 6 - 2
e^(1 + b) - e^b = 4
(e^1)(e^b) - e^b = 4
(e^b) (e - 1) = 4
e^b = 4/(e - 1)
b = ln(4/(e - 1))

Now that we have b, we can easily get B. Since
B = 6 - e^(1 + b)
B = 6 - e^(1 + ln(4/(e - 1)))

Conclusion: b = ln(4/(e - 1)) and B = 6 - e^(1 + ln(4/(e - 1))).

Answer 2

Plug it in, plug it in.
For (0,2) 2= e^b + B
and for (1,6) 6= e^(1+b) +B
From which (with some manipulations)
4 = e^(1+b) - e^b , which can be found by trial and error. Then B is easily found.
The values look reasonable, since if b=1, e^2 is about 7.3, and the difference is about 4.6.

Answer 3

y=e^(x+b) + B passess thro 2 points

2 = e^b + B >>>>e^b = 2 - B ---(1)

6 = e^(1+b) + B >>> e^1 * e^b = 6 -B ----(2)

(2) / (1) e^1 = 6 -B /2 - B
(2 - B )e^1 = 6 - B >>>
B = (2e^1 - 6) / (e^1-1) = - 0.564 / 1.718 = - 0.328 (roughly)

e^b = 2 - B = 2 + 0.328 = 2.328
take log
b = log (2.328) = 0.844 (roughly)

take more decimal places of e then will be exact

Answer 4

I'dotn know but thanks for the two points .. ps stop wasting your time on this web site be a teacher or something get a real job