How would you solve an exponential equation like this...
2^(3x+1)= 3^(x-2)
Take log on both sides to get:
(3x+1)log2 = (x-2)log3
Solve for x.
Log can be to any base.
2^(3x+1)= 3^(x-2)
(3x+1)log2 = (x-2)log3
Solve for x.
Log can be to any base.