How do you solve 5^(2x+1)=6^(x-2)?

multiply by log on both sides;
log(5^(2x+1)) = log(6^(x-2));
(2x+1)log 5 = (x-2)log 6;
(2x+1)/(x-2) = (log 6)/(log 5) = 1.1133;
2x+1 = 1.1133x - 2.2266;
0.8867x = -3.2266;
therefore;
x = (-3.2266)/0.8867 = 3.6389;
done.

5^(2x + 1) = 6^(x - 2)
log_5[5^(2x + 1)] = log_5[6^(x - 2)]
(2x + 1)log_5(5) = (x - 2)log_5(6)
2x + 1 = (x - 2)log_5(6)
(2x + 1)/(x - 2) = log_5(6)
(2x + 1)/(x - 2) ≈ 1.113
2x + 1 ≈ 1.113(x - 2)
2x + 1 ≈ 1.113x - 2.226
2x - 1.113x ≈ -1 - 2.226
0.887x ≈ -3.226
x ≈ -3.226/0.887
x ≈ -3.636

(2x + 1) log 5 = (x - 2) log 6

(2x) log 5 - (x) log 6 = - log 5 - 2 log 6

0.619 x = - ( log 5 + 2 log 6)

0.619 x = - 2.25

x = - 3.63

take ln both sides
(2x+1)ln5 = (x-2)ln6
solve for x, x= -3.6