How do you solve #lnx+ln(x+4)=ln5#?

Question

10667 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-10-31T19:04:03

Use the rule #log(a) + log(b) = log(a xx b)# to start the solving process.

#ln(x(x + 4)) = ln5#

#ln(x^2 + 4x) = ln5#

We can now use the property #loga = logb -> a = b# to solve.

#x^2 + 4x = 5#

#x^2 + 4x - 5 = 0#

#(x + 5)(x - 1) = 0#

#x = -5 and 1#

However, #x = -5# is extraneous, since the domain of #y = lnx# is #x >0#.

The only solution is hence #x = 1#.

Hopefully this helps!

Answer 2 · 2016-10-31T19:57:48

#ln(x)+ln(x+4)=ln(5)#

#ln(x(x+4))=ln(5)#

#x(x+4)=5#

#x^2+4x=5#

#x^2+4x-5=0#

Factorise:

#(x-1)(x+5)=0#

This means that:

x is equal to 1 as it can't be -5 . This is your answer.