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