How do I solve this equation?

Take log of both sides
So m log(1/9) = (m+4)log81
-m log9 = 2(m+4)log9......Using laws of logarithms.
So -m = 2m + 8 or m = - 8/3

(1/9)^m=81^m+4
Take log on both sides, we have
log(1/9)^m=log{81^(m+4)}
or m log1/9)=(m+4)log 81
m (log1- log9)=(m+4)log 9^2
or m (0 - log9)=2(m+4)log 9
or -m log9=2mlog 9 +8log9
or -m log9-2m log9 =8log9
-3m log9 =8log9
m = - 8/3

9^-m=9^(2m+8)
m=-8/3

(1/9)^m = 81^(m + 4)
(9^-1)^m = (9^2)^(m + 4)
9^(-m) = 9^[2(m + 4)]
log_9[9^(-m)] = log_9{9^[2(m + 4)]}
-m = 2(m + 4)
-m = 2m + 8
-m - 2m = 8
m = 8/-3
m = -8/3

ln(1/9)^m = ln(81)^m +ln4
using the rules of logs ln(a)^b = blna

m.ln(1/9)=m.ln(81) + ln4

m( ln(1/9) - ln(81) = ln 4

m = ln4 / ( ln(1/9) - ln(81) )

using rules of logs lna - lnb = ln(a/b)

m = ln4 / ln( (1/9)/81 )

first, subtract the (1/9)^m from the 81^m to get:
(728/9)^m=-4

Then, using log, we get:

log(base 728/9) -4 = m