What is the answer to 3^y+3^{1-y}=4 ?

The answer to 3^y+3^{1-y}=4 is y=1,y=0

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3^y+3^{1-y}=4

3^{y} + 3^{1 - y} = 4

y = 1, y = 0

Solution steps

3^{y} + 3^{1 - y} = 4

Apply exponent rules

3^{y} + (3^{y})^{- 1} \cdot 3^{1} = 4

Rewrite the equation with

3^{y} = u

u + (u)^{- 1} \cdot 3^{1} = 4

Solve

u + u^{- 1} \cdot 3^{1} = 4 : u = 3, u = 1

u = 3, u = 1

Substitute back

u = 3^{y},

solve for

y

Solve

3^{y} = 3 : y = 1

Solve

3^{y} = 1 : y = 0

y = 1, y = 0

7 = 3^{x}

3^{3 n} = 4^{3 n - y}

a, \frac{4 ^{a + 1}}{b} = 125

8^{x} = 32

2^{n} = 2000