# What is the solution of equation 6^(y+1)-5*6^y-1=0?

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The equation to be solved is 6^(y+1)-5*6^y-1=0

6^(y+1)-5*6^y-1=0

=> 6*6^y - 5*6^y - 1 = 0

=> 6^y - 1 = 0

=> 6^y = 1

=> y = 0

**The solution of the equation is y = 0**

The equation 6^(y+1)-5*6^y-1=0 has to be solved.

Use the relation `x^(a+b) = x^a*x^b`

This gives:

6^y*6^1 - 5*6^y - 1 = 0

6*6^y - 5*6^y - 1 = 0

6^y - 1 = 0

6^y = 1

As the result of any number raised to the power 0 is 1, y = 0

The solution of the equation is y = 0

We'll apply change the variable to solve the given exponential equation.

6^y = t

We'll express 6^(y+1)=(6^y)*6, based on the property of multiplying 2 exponential functions, having matching bases.

The equation will become:

6*6^y -5*6^y-1 = 0

But 6^y=t:

6t - 5t - 1 = 0

We'll combine like terms:

t - 1 = 0

We'll add 1 both sides:

t = 1

But 6^y = t=1

We could write 1=6^0

6^y=6^0

**The only possible solution of the given equation is y=0.**