Solution:

Given series: log 3 base 9 log 9 base 9 log 27 base 9 ... n terms

We can simplify the given terms using the following identities:

log a base a = 1 (for any positive number a)

log ab = log a + log b (product rule of logarithms)

Using these identities, we can simplify the given terms as follows:

log 3 base 9 = log 3 / log 9

log 9 base 9 = 1

log 27 base 9 = log 3^3 / log 9 = 3 log 3 / 2

Therefore, the given series becomes:

log 3 / log 9 + 1 + 3 log 3 / 2 log 9 + ...

Now, we need to find the sum of this infinite series.

To find the sum of an infinite geometric series, we use the formula:

Sum = a / (1 - r)

where a is the first term and r is the common ratio.

In this case, the first term is log 3 / log 9 and the common ratio is (3 log 3 / 2) / log 9.

Therefore, the sum of the given series is:

log 3 / log 9 / (1 - (3 log 3 / 2) / log 9)

Simplifying this expression, we get:

(log 3 / log 9) / (1 - (log 3)^3 / (2 log 9)^2)

which can be further simplified using the identity:

log a^b = b log a

This gives us:

(log 3 / log 9) / (1 - (log 3)^3 / (log 9)^3)

Now, we can substitute the values of log 3 and log 9:

log 3 = 0.4771

log 9 = 0.9542

Substituting these values, we get:

0.4771 / 0.0458 = 10.418

Therefore, the sum of the given series is 10.418.

Answer: 10.418