What is the answer to 8^{1+x}+8^{1-x}=20 ?

The answer to 8^{1+x}+8^{1-x}=20 is x= 1/3 ,x=-1/3

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8^{1+x}+8^{1-x}=20

8^{1 + x} + 8^{1 - x} = 20

x = \frac{1}{3}, x = - \frac{1}{3}

Decimal Notation

x = 0.33333 \dots, x = - 0.33333 \dots

Solution steps

8^{1 + x} + 8^{1 - x} = 20

Apply exponent rules

8^{x} \cdot 8^{1} + (8^{x})^{- 1} \cdot 8^{1} = 20

Rewrite the equation with

8^{x} = u

u \cdot 8^{1} + (u)^{- 1} \cdot 8^{1} = 20

Solve

u \cdot 8^{1} + u^{- 1} \cdot 8^{1} = 20 : u = 2, u = \frac{1}{2}

u = 2, u = \frac{1}{2}

Substitute back

u = 8^{x},

solve for

x

Solve

8^{x} = 2 : x = \frac{1}{3}

Solve

8^{x} = \frac{1}{2} : x = - \frac{1}{3}

x = \frac{1}{3}, x = - \frac{1}{3}

4^{x} + 2^{5} = 3 \cdot 2^{x + 2}

1 + 2 e^{x + 1} = 9

4^{x - 1} = 3

x^{7} = 7^{x}

e^{1 - 2 x} = e^{3 x - 5}

What is the answer to 8^{1+x}+8^{1-x}=20 ?
The answer to 8^{1+x}+8^{1-x}=20 is x= 1/3 ,x=-1/3
Solve for x: 8^{1+x}+8^{1-x}=20
The solution is x= 1/3 ,x=-1/3