What is the answer to e^{4x}-7e^{2x}+12=0 ?

The answer to e^{4x}-7e^{2x}+12=0 is x=ln(2),x= 1/2 ln(3)

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e^{4x}-7e^{2x}+12=0

e^{4 x} - 7 e^{2 x} + 12 = 0

x = ln (2), x = \frac{1}{2} ln (3)

Decimal Notation

x = 0.69314 \dots, x = 0.54930 \dots

Solution steps

e^{4 x} - 7 e^{2 x} + 12 = 0

Apply exponent rules

(e^{x})^{4} - 7 (e^{x})^{2} + 12 = 0

Rewrite the equation with

e^{x} = u

(u)^{4} - 7 (u)^{2} + 12 = 0

Solve

u^{4} - 7 u^{2} + 12 = 0 : u = 2, u = - 2, u = 3, u = - 3

u = 2, u = - 2, u = 3, u = - 3

Substitute back

u = e^{x},

solve for

x

Solve

e^{x} = 2 : x = ln (2)

Solve

e^{x} = - 2 :

No Solution for

x \in R

Solve

e^{x} = 3 : x = \frac{1}{2} ln (3)

Solve

e^{x} = - 3 :

No Solution for

x \in R

x = ln (2), x = \frac{1}{2} ln (3)

2^{- 3 d + 1} = 2^{- 2 d}

8^{4 x + 1} = 4^{3 x + 5}

5^{2 x + 5} = \frac{1}{5 ^{3 - x}}

- 14 (1 - 2^{x}) = 117638

t, x = e^{3 t}

What is the answer to e^{4x}-7e^{2x}+12=0 ?
The answer to e^{4x}-7e^{2x}+12=0 is x=ln(2),x= 1/2 ln(3)