in(x+5)=in(x-1)+in(x+1)

Question

find its equation with its solution.

Answer 1

ln(x+5) = ln(x-1) + ln(x+1)

 

ln(x+5) = ln[(x-1)(x+1)]

 

x+5 = x²-1

 

x²-x-6 = 0

 

(x-3)(x+2) = 0    x = 3 or -2

 

-2 is extraneous because it does not work in the original equation.  We can't find logarithms of negative numbers.  

 

The only solution is 3.  

 

Answer 2

If you do mean ln, then your first step to do is to first condense the logarithm on the right hand side of the equation. Remember that a logarithm is just an exponent. when you add exponents that means they have the same base and you just multiply.

 

This gives us ln(x+5)=ln((x-1)(x+1)). since both equations have ln, then the equations inside must be equal for x to have a solution.

 

So x+5=(x-1)(x+1), x+5=x²-1, as soon as you see x², you should notice that is a quadratic equation so, set it equal to zero to solve by factoring or quadratic formula. We get x²-x-6=0, factor and we get (x-3)(x+2)=0,this means x=-2, and x=3. With logarithmic equations always plug in your solutions to check for extraneous solutions into the original equation.  Remember we only want real number solutions so, we are not allowed to take the ln of a negative number so, we must reject -2. So, the only solution is x=3