Just a Simple Logarithm

$\log_{a}{b} + (\log_{a}{b})^{2} + (\log_{a}{b})^{3} + ........ = 2$ $\log_{a}{b} + \log_{b}{n} = ....$ $n = a^{\frac{2}{3}}$

1. First let $x = \log_{a}{b}$ , and change the equation accordingly

$x + x^{2} + x^{3} + ........ = 2$

1. Next we solve for $x$ .

$x^{2} + x^{3} + x^{4} + ........ = 2x$ $2 - 2x = x$ $2 = 3x$ $x = \frac{2}{3}$

1. We now know that $\log_{a}{b} = \frac{2}{3}$ , this also implies the following

$n = a^{\frac{2}{3}} = b$ $n = b$

1. We then substitute both values into the final equation to get

$\frac{2}{3} + \log_{b}{b}$ $\frac{2}{3} + 1$ $\boxed{\frac{5}{3}}$

The equation :

$log_b^a + (log_b^a)^2 + (log_b^a)^3 + . . . . . . . = 2$ is an geometric progression with $log_b^a$ as the first term and the ratio. Thus, we can input it to the geometric progression's sum formula :

$2 = \frac{log_b^a}{1-log_b^a}$

$2(log_a^a - log_b^a) = log_b^a$

$2 log_{\frac{a}{b}}^a = log_b^a$

Thus,

$\frac{a^2}{b^2} = b$

$a^2 = b^3$ => $a = b^{\frac{3}{2}}$ or $b = a^{\frac{2}{3}}$

Then, $log_b^a = \frac{2}{3} log_a^a\ = \frac{2}{3}$

and $log_n^b = log_b^b = 1$

$log_b^a + log_n^b = \frac{2}{3} + 1 = \boxed{\frac{5}{3}}$