What is the answer to 7^{2x}-8*7^x+7=0 ?

The answer to 7^{2x}-8*7^x+7=0 is x=1,x=0

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7^{2x}-8*7^x+7=0

7^{2 x} - 8 \cdot 7^{x} + 7 = 0

x = 1, x = 0

Solution steps

7^{2 x} - 8 \cdot 7^{x} + 7 = 0

Apply exponent rules

(7^{x})^{2} - 8 \cdot 7^{x} + 7 = 0

Rewrite the equation with

7^{x} = u

(u)^{2} - 8 u + 7 = 0

Solve

u^{2} - 8 u + 7 = 0 : u = 7, u = 1

u = 7, u = 1

Substitute back

u = 7^{x},

solve for

x

Solve

7^{x} = 7 : x = 1

Solve

7^{x} = 1 : x = 0

x = 1, x = 0

5^{x} = 3^{x + 2}

8^{n} - 8^{n - 1} = 14

4^{x} = 0

4^{x} = 6

e^{2 x} - 7 e^{x} = - 10

What is the answer to 7^{2x}-8*7^x+7=0 ?
The answer to 7^{2x}-8*7^x+7=0 is x=1,x=0
Solve for x: 7^{2x}-8*7^x+7=0
The solution is x=1,x=0