What is the real solution of equation 5^(t+1)=4*5^t+1

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justaguide | College Teacher | (Level 2) Distinguished Educator

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We have to solve the equation: 5^(t+1) = 4*5^t + 1

5^(t+1) = 4*5^t + 1

=> 5^t * 5 = 4* 5^t + 1

let y = 5^t

=> y*5 = 4*y + 1

=>5y - 4y = 1

=> y = 1

Now 5^t = y = 1

=> t = 0 , as any number raised to the power 0 is 1.

Therefore, we get the solution t = 0 .

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

We'll apply substitution technique to solve the given exponential equation.

5^t = y

The next step would be to express 5^(t+1)=(5^t)*5, based on the property of multiplying 2 exponential functions, having matching bases. The result of multiplication will be the base raised to the sum of exponents of each exponential function.

The equation will become:

5*5^t -4*5^t-1 = 0

But 5^t=y:

5y - 4y - 1 = 0

We'll combine like terms:

y - 1 = 0

We'll add 1 both sides:

y = 1

But 5^t = y=1

We could write 1=5^0

5^t=5^0

The real solution of the given equation is t=0.

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