Geometric sequence help please

ok so 1st problem i need to find n
a=2(a is first term)
r=3 (ratio)
a_n = 118,098
answer is n =11 but i need the steps

so here is my work
2(3)^(n-1) = 118,098
3^(n-1) = 59,049
what is the next step?

alrite next problem
find the sum
1 - 1/2 + 1/4 - 1/8 + ... - 1/512

i duno how to start and dont know the answeer

please help

so here is my work
2(3)^(n-1) = 118,098
3^(n-1) = 59,049
what is the next step?
(n-1)log3 = log59049
n-1 = 10
n =11
For the next one a = 1 and r = -1/2
You need to find the sum up to term -1/512
Find n by the same method as above.
Then the formula for the sum of a geometric sequence is
a(1-r^n)/(1-r)

Use the power of a logarithm property.

so log 3^(n-1) = log 59049
(n-1) log 3 = log 59049

then divide log 59049 by log 3, move thte one, and you have n!