If Y gas a geometric distribution with success probability 0.3,what is the largest value of y such that P(Y>y)≥0.1

Question

I had my answer to be 6.But I'm not sure

Answer 1

P=0.3
q=1-p =0.7
P(Y=y)=q^y=0.7^y

y such that 0.7^y=0.1
I apply ln to both sides and I get y to be 6

But I'm not sure with my answer