What is the answer to ln(x^2+3)-ln(x^2)=1 ?

The answer to ln(x^2+3)-ln(x^2)=1 is x=sqrt(3/(e-1)),x=-sqrt(3/(e-1))

Solve for x: ln(x^2+3)-ln(x^2)=1

The solution is x=sqrt(3/(e-1)),x=-sqrt(3/(e-1))

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ln (x^{2} + 3) - ln (x^{2}) = 1

x = \frac{3}{e - 1}, x = - \frac{3}{e - 1}

Decimal Notation

x = 1.32133 \dots, x = - 1.32133 \dots

Solution steps

ln (x^{2} + 3) - ln (x^{2}) = 1

Add

ln (x^{2})

to both sides

ln (x^{2} + 3) - ln (x^{2}) + ln (x^{2}) = 1 + ln (x^{2})

Simplify

ln (x^{2} + 3) = 1 + ln (x^{2})

Apply log rules

x^{2} + 3 = e x^{2}

Solve

x^{2} + 3 = e x^{2} : x = \frac{3}{e - 1}, x = - \frac{3}{e - 1}

x = \frac{3}{e - 1}, x = - \frac{3}{e - 1}

Verify Solutions:

x = \frac{3}{e - 1}

True

, x = - \frac{3}{e - 1}

True

The solutions are

x = \frac{3}{e - 1}, x = - \frac{3}{e - 1}

ln (x) ln (x) = 1

lo g_{x} (3 x + 4) = lo g_{x} (4 x + 2)

lo g_{4} (x) - lo g_{x} (64) = 2

lo g_{3} (x + 2) + lo g_{3} (3 x) = 2

- \frac{ln ( p )}{ln ( 7 )} - 4 = 0

What is the answer to ln(x^2+3)-ln(x^2)=1 ?
The answer to ln(x^2+3)-ln(x^2)=1 is x=sqrt(3/(e-1)),x=-sqrt(3/(e-1))
Solve for x: ln(x^2+3)-ln(x^2)=1
The solution is x=sqrt(3/(e-1)),x=-sqrt(3/(e-1))