What is the answer to 7^{2x+1}-2\times 7^{x+1}+7=0 ?

The answer to 7^{2x+1}-2\times 7^{x+1}+7=0 is x=0

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7^{2x+1}-2\times 7^{x+1}+7=0

7^{2 x + 1} - 2 \cdot 7^{x + 1} + 7 = 0

x = 0

Solution steps

7^{2 x + 1} - 2 \cdot 7^{x + 1} + 7 = 0

Apply exponent rules

(7^{x})^{2} \cdot 7^{1} - 2 \cdot 7^{x} \cdot 7^{1} + 7 = 0

Rewrite the equation with

7^{x} = u

(u)^{2} \cdot 7^{1} - 2 u \cdot 7^{1} + 7 = 0

Solve

u^{2} \cdot 7^{1} - 2 u \cdot 7^{1} + 7 = 0 : u = 1

u = 1

Substitute back

u = 7^{x},

solve for

x

Solve

7^{x} = 1 : x = 0

x = 0

3^{1 - y} - 3^{y} = 2

4 e^{x} = 0

6 e^{- x} x^{5} - e^{- x} x^{6} = 0

500 e^{- x} = 300

What is the answer to 7^{2x+1}-2\times 7^{x+1}+7=0 ?
The answer to 7^{2x+1}-2\times 7^{x+1}+7=0 is x=0
Solve for x: 7^{2x+1}-2\times 7^{x+1}+7=0
The solution is x=0