9^x = 2e^(x^2) — step-by-step solution

1) Let's solve in

ℝ

9^{x} = 2 e^{x^{2}}

You can apply the ln function on both sides:

\ln (9^{x}) = \ln (2 e^{x^{2}})

ⓘ On the real domain of the original expression, the base is strictly positive: the logarithm of the power is the exponent times the logarithm of the base:

\ln (9^{x})

x \cdot \ln (9)

■On the real domain of the original expression, the factors are strictly positive: the logarithm of the product is the sum of the logarithms, that of the quotient their difference:

\ln (2 e^{x^{2}})

\ln (2) + \ln (e^{x^{2}})

x \cdot \ln (9) = \ln (2) + \ln (e^{x^{2}})

ⓘ The logarithm of a power is a multiple of the logarithm, ln(a^n) = n*ln(a):

\ln (9)

\ln (3^{2})

2 \cdot \ln (3)

■For real

x

\ln (\exp (x)) = x

; wherever

\ln (x)

is defined,

\exp (\ln (x)) = x

\ln (e^{x^{2}})

x^{2}

x \cdot 2 \cdot \ln (3) = \ln (2) + x^{2}

- x^{2} + 2 x \cdot \ln (3) - \ln (2) = 0

Since

-1 \neq 0

, divide both sides by

-1

x^{2} - 2 x \cdot \ln (3) + \ln (2) = 0

Given the quadratic form

a x^{2} + b x + c

, let's compute the discriminant:

Δ = b^{2} - 4 a c

. Here

a = 1

b = -2 \cdot \ln (3)

and

c = \ln (2)

Δ = {(-2 \cdot \ln (3))}^{2} - 4 \cdot 1 \cdot \ln (2)

ⓘ Powers are computed before multiplications, additions and subtractions. With an even integer exponent, the minus sign in the base disappears; with an odd integer exponent, it moves in front of the power:

{(- 2 \cdot \ln (3))}^{2}

{(2 \cdot \ln (3))}^{2}

Δ = {(2 \cdot \ln (3))}^{2} - 4 \cdot \ln (2)

ⓘ The power of a product is the product of the powers:

{(2 \cdot \ln (3))}^{2}

2^{2} \cdot {\ln (3)}^{2}

Δ = 2^{2} \cdot {\ln (3)}^{2} - 4 \cdot \ln (2)

ⓘ Powers are computed before multiplications, additions and subtractions. An integer power is a repeated product:

2^{2}

2 \cdot 2

4

Δ = 4 \cdot {\ln (3)}^{2} - 4 \cdot \ln (2)

Since

Δ

\approx 2.0552 > 0

, the equation has two roots:

The quadratic formula gives

x_{1}

\frac{- b - \sqrt{Δ}}{2 a}

\frac{- (-2 \cdot \ln (3)) - \sqrt{4 \cdot {\ln (3)}^{2} - 4 \cdot \ln (2)}}{2 \cdot 1}

x_{1} = \frac{- (-2 \cdot \ln (3)) - \sqrt{4 \cdot {\ln (3)}^{2} - 4 \cdot \ln (2)}}{2}

ⓘ Extract the minus signs: an even number of minus signs cancels out, an odd number leaves one minus sign:

- (-2 \cdot \ln (3))

2 \cdot \ln (3)

■A factor that is an exact power comes out of the root:

\sqrt{4 \cdot {\ln (3)}^{2} - 4 \cdot \ln (2)}

2 \cdot \sqrt{{\ln (3)}^{2} - \ln (2)}

x_{1} = \frac{2 \cdot \ln (3) - 2 \cdot \sqrt{{\ln (3)}^{2} - \ln (2)}}{2}

ⓘ A fraction is simplified by dividing the numerator and the denominator by the same number:

\frac{2 \cdot \ln (3) - 2 \cdot \sqrt{{\ln (3)}^{2} - \ln (2)}}{2}

\frac{2 \cdot (\ln (3) - \sqrt{{\ln (3)}^{2} - \ln (2)})}{2}

\ln (3) - \sqrt{{\ln (3)}^{2} - \ln (2)}

x_{1} = \ln (3) - \sqrt{{\ln (3)}^{2} - \ln (2)}

The quadratic formula gives

x_{2}

\frac{- b + \sqrt{Δ}}{2 a}

\frac{- (-2 \cdot \ln (3)) + \sqrt{4 \cdot {\ln (3)}^{2} - 4 \cdot \ln (2)}}{2 \cdot 1}

x_{2} = \frac{- (-2 \cdot \ln (3)) + \sqrt{4 \cdot {\ln (3)}^{2} - 4 \cdot \ln (2)}}{2}

ⓘ Extract the minus signs: an even number of minus signs cancels out, an odd number leaves one minus sign:

- (-2 \cdot \ln (3))

2 \cdot \ln (3)

■A factor that is an exact power comes out of the root:

\sqrt{4 \cdot {\ln (3)}^{2} - 4 \cdot \ln (2)}

2 \cdot \sqrt{{\ln (3)}^{2} - \ln (2)}

x_{2} = \frac{2 \cdot \ln (3) + 2 \cdot \sqrt{{\ln (3)}^{2} - \ln (2)}}{2}

ⓘ A fraction is simplified by dividing the numerator and the denominator by the same number:

\frac{2 \cdot \ln (3) + 2 \cdot \sqrt{{\ln (3)}^{2} - \ln (2)}}{2}

\frac{2 \cdot (\ln (3) + \sqrt{{\ln (3)}^{2} - \ln (2)})}{2}

\ln (3) + \sqrt{{\ln (3)}^{2} - \ln (2)}

x_{2} = \ln (3) + \sqrt{{\ln (3)}^{2} - \ln (2)}

▶Solutions obtained:

x = {\ln (3) - \sqrt{{\ln (3)}^{2} - \ln (2)}, \ln (3) + \sqrt{{\ln (3)}^{2} - \ln (2)}}

2) The numeric value of:

x = {\ln (3) - \sqrt{{\ln (3)}^{2} - \ln (2)}, \ln (3) + \sqrt{{\ln (3)}^{2} - \ln (2)}}

equals:

x = {\approx 1.0986 - {({1.0986}^{2} - 0.6931)}^{0.5}, \approx 1.0986 + {({1.0986}^{2} - 0.6931)}^{0.5}}

x = {\approx 1.0986 - {(1.2069 - 0.6931)}^{0.5}, \approx 1.0986 + {(1.2069 - 0.6931)}^{0.5}}

x = {\approx 1.0986 - {0.5138}^{0.5}, \approx 1.0986 + {0.5138}^{0.5}}

x = {\approx 1.0986 - 0.7168, \approx 1.0986 + 0.7168}

▷

x = {\approx 0.3818, \approx 1.8154}