What is the answer to 7(7^{2x})-50(7^x)+7=0 ?

The answer to 7(7^{2x})-50(7^x)+7=0 is x=1,x=-1

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7(7^{2x})-50(7^x)+7=0

7 (7^{2 x}) - 50 (7^{x}) + 7 = 0

x = 1, x = - 1

Solution steps

7 (7^{2 x}) - 50 (7^{x}) + 7 = 0

Apply exponent rules

(7^{x})^{2} \cdot 7^{1} - 50 \cdot 7^{x} + 7 = 0

Rewrite the equation with

7^{x} = u

(u)^{2} \cdot 7^{1} - 50 u + 7 = 0

Solve

u^{2} \cdot 7^{1} - 50 u + 7 = 0 : u = 7, u = \frac{1}{7}

u = 7, u = \frac{1}{7}

Substitute back

u = 7^{x},

solve for

x

Solve

7^{x} = 7 : x = 1

Solve

7^{x} = \frac{1}{7} : x = - 1

x = 1, x = - 1

93 = 3 1^{x}

- e^{x} = e^{- x}

2^{5 x - 1} = 3

3^{y} = 1

7^{x} = 3^{x + 1}

What is the answer to 7(7^{2x})-50(7^x)+7=0 ?
The answer to 7(7^{2x})-50(7^x)+7=0 is x=1,x=-1
Solve for x: 7(7^{2x})-50(7^x)+7=0
The solution is x=1,x=-1