Factor b:
\(b(\frac{1}{7}+\frac{b}{7^2}+\frac{b^2}{7^3}+...)=1\)
Now, what is inside the bracket is a geometric series with a common ratio, \(r=\frac{b}{7}\)
The formula is:
\(Sum=\frac{1}{1-r}=\frac{1}{1-\frac{b}{7}}=\frac{7}{7-b}\) (But, r has to be less than 1).
So:
\(b(\frac{7}{7-b})=1\), can you solve it from here?