Find the value of b for which 1+e^b+e^2b+e^3b+...=9

this is a geometric series, let r=e^b then we have
1+r+r^2+r^3+....=9
1/(1-r)=9
1-r=(1/9)
r=1-(1/9)
r=8/9
e^b=8/9 take ln of both sides
ln(e^b)=8/9 ln(x^y)=y*ln(x) and ln(e)=1 so
b*ln(e)=b=ln(8/9)
thus
b=ln(8/9)

Answer Link

astha8579 astha8579 · Answer 2 · 2022-01-07T07:22:34+01:00

The value of b is:

[tex]b = ln(\dfrac{8}{9})[/tex]

The series that you've got is called Geometric Series in mathematics.

A Geometric Series is one in which each subsequent term is a constant multiple times more than that of previous term.

Its standard form is:

[tex]S_n = a + ar + ar^2 + ar^3 + \cdots + ar^{n-1}[/tex]

The sum of such series with infinite terms is given as:

[tex]S_{\inf} = \dfrac{a}{1-r}; \: (|r| < 1)\\[/tex]

Sum of given series, assuming |[tex]e^b[/tex]| < 1 is:

[tex]1+e^b+e^{2b}+e^{3b}+...\\\\= \dfrac{1}{1-e^b}[/tex]

or we have:

[tex]\dfrac{1}{1-e^b} = 9\\\\or\\\\9e^b = 9 - 1\\\\\\b = ln(\dfrac{8}{9})[/tex]

Thus we have :

[tex]b = ln(\dfrac{8}{9})[/tex]

Learn more here:

https://brainly.com/question/3924955