Solve the equation 25^(square root(x+1))+5=6*5^squareroot(x+1)

We notice that 25^sqrt(x+1) =
5^2sqrt(x+1)

We can replace 5^sqrt(x+1) by
t:

The equation will
become:

t^2 + 5 = 6t

We'll
shift all terms to the left:

t^2 - 6t + 5 =
0

We'll apply quadratic
formula:

t1 =
[6+sqrt(36-20)]/2

t1 =
(6+sqrt16)/2

t1 =(6+4)/2

t1 =
5

t2 = 1

But t1 = 5^sqrt(x+1)
=> 5 = 5^sqrt(x+1)

Since the bases are matching,
we'll apply one to one rule:
sqrt(x+1) = 1

We'll
raise to square both sides:

x + 1 = 1 => x =
0

t2 = 5^sqrt(x+1) => 1 = 5^sqrt(x+1), where 1 =
5^0

Since the bases are matching, we'll apply again one to
one rule:
sqrt(x+1) = 0

x + 1 = 0 => x =
-1

Both value are valid, therefore the
solutions of the equation are {-1 ; 0}.