How do you solve #log(x+3)+log(x-3)=log27#?

1 Answer

Answer:

#x = 6#

Explanation:

First of all, this equation is defined on #]3,+oo[# because you need #x+3 > 0# and #x - 3 > 0# at the same time or the log won't be defined.

The log function maps a sum into a product, hence #log(x+3) + log(x-3) = 27 iff log [(x+3)(x-3)] = log 27#.

You now apply the exponential function on both sides of the equation : #log [(x+3)(x-3)] = log 27 iff (x+3)(x-3) = 27 iff x^2 - 9 = 27 iff x^2 - 36 = 30#. This is a quadratic equation that has 2 real roots because #Delta = -4*(-36) = 144 > 0#

You know apply the quadratic formula #x = (-b+- sqrtDelta)/2a# with #a = 1# and #b = 0#, hence the 2 solutions of this equation : #x = ± 6#

#-6 !in ]3,+oo[# so we can't keep this one. The only solution is #x = 6#.