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

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We have to solve 25^(sqrt (x+1)) + 5 = 6*5^sqrt(x+1)

25^(sqrt (x+1)) + 5 = 6*5^sqrt(x+1)

=> (5^2)^(sqrt (x+1)) + 5 = 6*5^sqrt(x+1)

=> 5^2(sqrt (x+1) + 5 = 6*5^sqrt(x+1)

Let 5^(sqrt (x + 1)) = y

=> y^2 + 5 = 6y

=> y^2 - 6y + 5 = 0

=> y^2 - 5y - y + 5 = 0

=> (y - 5) - 1(y - 5) = 0

=> (y - 5)(y - 1) = 0

=> y = 5 and y = 1

For y = 5, 5^(sqrt (x + 1)) = 5

=> sqrt (x + 1) = 1

=> x + 1 = 1

=> x = 0

For y = 1, 5^(sqrt (x + 1)) = 1

=> sqrt (x + 1) = 0

=> x + 1 = 0

=> x = -1

**The roots of the given equation are x = 0 and 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}.**