Algebra
posted by
Anna
.
How do I start to solve 2log8(x+1) = 8?
I think 2log8(x+1) is log8^(x+1)to the 2nd but then what do I do because then I think I should do log8/log8 = x+1)^2 but that doesn't work at all
Thank you for any help you can give

AlgebraCould Mathmate or Reiny or Dr. Bob look at this if you have time 
Anna
Please give me some direction for the above problem
Thank you

Algebra 
Reiny
2log8(x+1) = 8
first of all, multiply be 2
log8(x+1) = 4
log_{8}(x+1)^2 = 8
(x+1) = 8^4
x = 4095