What is the answer to 7^{2x}+7^x-12=0 ?

The answer to 7^{2x}+7^x-12=0 is x=(ln(3))/(ln(7))

Solve for x: 7^{2x}+7^x-12=0

The solution is x=(ln(3))/(ln(7))

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7^{2 x} + 7^{x} - 12 = 0

x = \frac{ln ( 3 )}{ln ( 7 )}

Decimal Notation

x = 0.56457 \dots

Solution steps

7^{2 x} + 7^{x} - 12 = 0

Apply exponent rules

(7^{x})^{2} + 7^{x} - 12 = 0

Rewrite the equation with

7^{x} = u

(u)^{2} + u - 12 = 0

Solve

u^{2} + u - 12 = 0 : u = 3, u = - 4

u = 3, u = - 4

Substitute back

u = 7^{x},

solve for

x

Solve

7^{x} = 3 : x = \frac{ln ( 3 )}{ln ( 7 )}

Solve

7^{x} = - 4 :

No Solution for

x \in R

x = \frac{ln ( 3 )}{ln ( 7 )}

y, y = e^{x + y}

2^{4 x + 1} \times 8^{- x + 3} = \frac{1}{16}

2^{x + 1} + 2^{x} + 2^{x - 1} = 7

e^{\frac{x}{2}} = 2 e^{2 x}

e^{x + 1} = 6^{x + 6}

What is the answer to 7^{2x}+7^x-12=0 ?
The answer to 7^{2x}+7^x-12=0 is x=(ln(3))/(ln(7))
Solve for x: 7^{2x}+7^x-12=0
The solution is x=(ln(3))/(ln(7))