What is the answer to e^{(2t+1)}=9e^{(1-t)} ?

The answer to e^{(2t+1)}=9e^{(1-t)} is t=(2ln(3))/3

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e^{(2t+1)}=9e^{(1-t)}

e^{(2 t + 1)} = 9 e^{(1 - t)}

t = \frac{2 ln ( 3 )}{3}

Decimal Notation

t = 0.73240 \dots

Solution steps

e^{(2 t + 1)} = 9 e^{(1 - t)}

Apply exponent rules

2 t + 1 = ln (9) + 1 - t

Solve

2 t + 1 = ln (9) + 1 - t : t = \frac{2 ln ( 3 )}{3}

t = \frac{2 ln ( 3 )}{3}

x, 2 e^{3 x} = 4 e^{5 x}

5^{x - 3} = 10

x, 2^{x} - 2^{y} = 128

6 = 12 \cdot 5^{- k^{20}}

7^{- x} \cdot 7^{- x} = 343

What is the answer to e^{(2t+1)}=9e^{(1-t)} ?
The answer to e^{(2t+1)}=9e^{(1-t)} is t=(2ln(3))/3