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

Question

I just don't even know what to do with this; solve for x: ln(x^2+1)-3lnx=ln(2)

Answer 1

When we subtract logs of common bases, we can divide the arguments of the logs.  Also, the coefficient of the log becomes of the exponent of the argument.

 

ln((x² + 1) / x³) = ln(2)

 

 

Equate the arguments.

 

(x² + 1) / x³ = 2

 

 

Multiply both sides of the equation by x³.

 

x² + 1 = 2x³

 

 

Subtract x² and 1 on both sides of the equation.

 

0 = 2x³ - x² - 1

 

 

Using synthetic division to factor the right side of the equation,

 

0 = (x - 1)(2x² + x + 1)

 

 

Set the factors equal to zero.

 

x - 1 = 0          and          2x² + x + 1 = 0

 

x = 1

 

 

To find the roots of the other factor, we can use the quadratic formula, or we can calculate the discriminant to see whether we obtain real or complex roots.  The discriminant is the square-root part of the formula.

 

b² - 4ac = 1² - 4(2)(1) = 1 - 8 = -7

 

This tells us that this factor does not have any real roots.

 

 

Your solution is

 

x = 1

Answer 2

How about starting from the properties of logs:

 

a*ln(b) = ln(a^b)

ln(a)-ln(b) = ln(a/b)

 

Also, recall that ln(x) is defined only if x > 0. 

 

So ln(x^2+1) - 3ln(x)  = ln(x^2+1) - ln(x^3) = ln(x^2+1/x^3) = ln(2)

 

......

 

Now it is easy to get rid of the natural logs and you have a normal algebraic equation.  Remember that your solutions require that x be positive and that if you end up with negative solutions or complex solutions, you will have to discard them. 

 

Answer 3

use divsion property of logs on left side of equation

 

ln((x^2+1))/3)=ln(2)

(X^2+1)/3=2

solve for x

remember you cannot take the log of 0 or a negative number!!