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

2 Answers
Oct 31, 2016

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!

Oct 31, 2016

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

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

See logarithmic rules .

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.