What is the answer to (ln(x))^2-ln(x^7)+12=0 ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(ln(x))^2-ln(x^7)+12=0

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

x = e^{4}, x = e^{3}

Decimal Notation

x = 54.59815 \dots, x = 20.08553 \dots

Solution steps

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

Apply log rules

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

Rewrite the equation with

ln (x) = u

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

Solve

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

u = 4, u = 3

Substitute back

u = ln (x),

solve for

x

Solve

ln (x) = 4 : x = e^{4}

Solve

ln (x) = 3 : x = e^{3}

x = e^{4}, x = e^{3}

Verify Solutions:

x = e^{4}

True

, x = e^{3}

True

The solutions are

x = e^{4}, x = e^{3}

lo g_{7} (x + 4) - lo g_{7} (x) = 3

lo g_{2} ((r + 10)^{2}) = 4

lo g_{2} (x - 6) = 5

lo g_{2} ((r + 6)^{2}) = 6

lo g_{64} (x) = \frac{3}{2}

What is the answer to (ln(x))^2-ln(x^7)+12=0 ?
The answer to (ln(x))^2-ln(x^7)+12=0 is x=e^4,x=e^3
Solve for x: (ln(x))^2-ln(x^7)+12=0
The solution is x=e^4,x=e^3