How do you solve lnx+ln(x-2)=1?

2 Answers
Gió
Jul 20, 2016

I found: x=1+sqrt(1+e)

Explanation:

We can use the property of logs that allows us to change the sum into a product of arguments as:
ln[x*(x-2)]=1
then use the definition of log, applied to our natural log, to write:
x(x-2)=e^1
x^2-2x-e=0
solve for x to get:
x_(1,2)=(2+-sqrt(4+4e))/2=(2+-2sqrt(1+e))/2
x_1=1+sqrt(1+e)
x_2=1-sqrt(1+e) NOT (because negative)

Costas C.
Jul 20, 2016

Use the properties:

lna+lnb=ln(a*b)

lnx=y=>x=e^y

Then solve the quadratic equation. Answer is:

x_1=2.928

x_2=-0.928

Explanation:

lnx+ln(x-2)=1

Since lna+lnb=ln(a*b)

ln(x*(x-2))=1

To add an ln function to the right side we use x=lne^x

ln(x*(x-2))=lne^1

lnx is a 1-1 function:

x*(x-2)=e

x^2-2x-e=0

Solving the quadratic:

Δ=(-2)^2-4*1*(-e)=4+4e

x_(1,2)=(-(-2)+-sqrt(4+4e))/(2*1)

From the calculator sqrt(4+4e)=3.857

x_1=2.928

x_2=-0.928

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