Solve for real t 11^(t+1) -1=4*11^t
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We have to solve 11^(t+1) -1=4*11^t for t.
11^(t+1) -1=4*11^t
=> 11^t*11 - 1 = 4*11^t
let 11^t = x
=> 11x - 1 = 4x
=> 7x = 1
=> x= 1/7
11^t = 1/7
Here the bases are different, so we have to use the log function.
log 11^t = log (1/7)
=> t* log 11 = -log 7
=> t = -log 7 / log 11
The required value of t is -log 7 / log 11
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We'll apply substitution technique to solve the given exponential equation.
11^t = y
We'llÂ express 11^(t+1)=(11^t)*11, 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.
We'll move all terms to one side and we'll get:
11*11^t -4*11^t-1 = 0
But 11^t=y:
11y - 4y - 1 = 0
We'll combine like terms:
7y - 1 = 0
We'll add 1 both sides:
7y = 1
y = 1/7
But 11^t = y=1/7
11^t = 1/7
We'll take logarithms both sides:
ln (11^t) = ln (1/7)
t*ln11 = ln (1/7)
t = ln (1/7)/ln11
The real solution of the given equation is t=ln (1/7)/ln11.
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