How do you solve #log(x)+log(x+1)=log(12)#?

1 Answer
Topscooter
Dec 17, 2015

The answer is #x = 3#.

Explanation:

You first have to say where the equation is defined : it is defined if #x > -1# since the logarithm can't have negative numbers as argument.

Now that this is clear, you now have to use the fact that natural logarithm maps addition into multiplication, hence this :

#ln(x) + ln(x + 1) = ln(12) iff ln[x(x + 1)] = ln(12)#

You can now use the exponential function to get rid of the logarithms :

#ln[x(x + 1)] = ln(12) iff x(x+1) = 12#

You develop the polynomial at the left, you substract 12 at both sides, and you now have to solve a quadratic equation :

#x(x+1) = 12 iff x^2 + x - 12 = 0#

You now have to calculate #Delta = b^2 - 4ac#, which here equals to #49# so this quadratic equations has two real solutions, given by the quadratic formula : #(-b+sqrt(Delta))/(2a)# and #(-b-sqrt(Delta))/(2a)#. The two solutions here are #3# and #-4#. But the very 1st equation we are solving right now is only defined for #x > -1# so #-4# is not a solution of our log equation.

Related questions
See all questions in Logarithmic Models
Impact of this question
9331 views around the world
You can reuse this answer
Creative Commons License