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

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Answer 1 · 2018-07-21T15:19:03

#color(blue)(x=(-1+sqrt(4e+9))/2~~1.7290)#

Using the logarithmic property:

#lna+lnb=lnab#

#ln(x-1)(x+2)=1#

Using the antilogarithm:

#e^(ln(x-1)(x+2))=e^1#

#(x-1)(x+2)=e#

Expanding #LHS#

#x^2+x-2-e=0#

Using quadratic formula:

#x=(-(1)+-sqrt((1)^2-4(1)(-2-e)))/2#

#x=(-1+sqrt(4e+9))/2~~1.7290#

#x=(-1-sqrt(4e+9))/2~~-2.7290#

Checking solutions with original equation:

#ln(1.7290-1)+ln(1.7290+2)=1.000058554#

#ln(-2.7290-1)+ln(-2.7290+2) \ \ color(red)(X)#

This is undefined for real numbers.

#lnx# is only defined for real numbers if #x>0#

Answer 2 · 2018-07-21T15:33:55

#x=\frac{-1+\sqrt{9+4e}}{2}=1.72896#

#\ln(x-1)+\ln(x+2)=1\ quad (\forall \ \ x>1)#

#\ln((x-1)(x+2))=1#

#\ln(x^2+x-2)=\ln e#

Comparing the numbers on same base on both the sides,

#x^2+x-2=e#

#x^2+x-(2+e)=0#

Solving above quadratic equation as follows

#x=\frac{-1\pm\sqrt{1^2-4(1)(-(2+e))}}{2(1)}#

#x=\frac{-1\pm\sqrt{9+4e}}{2}#

But, #x>1# hence

#x=\frac{-1+\sqrt{9+4e}}{2}#

#=1.72896#