What is the answer to log_{3}(2^x+1)=2 ?

The answer to log_{3}(2^x+1)=2 is x=3

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log_{3}(2^x+1)=2

lo g_{3} (2^{x} + 1) = 2

x = 3

Solution steps

lo g_{3} (2^{x} + 1) = 2

Rewrite the equation with

2^{x} = u

lo g_{3} (u + 1) = 2

Solve

lo g_{3} (u + 1) = 2 : u = 8

u = 8

Substitute back

u = 2^{x},

solve for

x

Solve

2^{x} = 8 : x = 3

x = 3

1 = (3)^{x}

[4^{x + 1}] - (4^{x}) = 48

e^{- 2 x} - 2 e^{- x} = 0

e^{x} = - 6

2^{n - 3} \cdot 1 4^{4 n - 5} = \frac{1}{56}