need help

5^b + 5^b + 5^b + 5^b + 5^b = 25^(b - 1)

 

5 * 5^b ==25^(b - 1)

 

5^1  *  5^b ==25^(b  -  1)

 

5^(b + 1) == 25^(b  -  1)

 

log(5)b  +  log(5) ==log(25)b  -  log(25)

 

(log(5)  -  log(25))b == - log(5)  -  log(25), solve for b

 

b == [- log(5) - log(25)] / [log(5)  -  log(25)]

 

b ==3

A non-log solution.

5^b + 5^b + 5^b + 5^b + 5^b = 25^(b - 1),

5.5^b = 5^(b - 1).5^(b - 1),

          = 5^(b - 1).5^b / 5,

          = 5^(b - 2). 5^b,

1 = b - 2,

b = 3.

Yet another approach:

 

5*5^b = (5^2)^(b-1)

 

5^(b+1) = 5^(2*(b-1))

 

So b+1 = 2*(b-1)  or  b = 3