# What is the farthist this equation can go? ..... 2ln(x-1)= 2ln(x-2) Its precalculus related

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Use the property of logarithms: `a*ln b= ln(b^a)`

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

`2ln(x-2)=ln(x-2)^2`

`ln(x-1)^2=ln(x-2)^2`

The function logarithm is bijective, meaning that if `ln(x-1)^2=ln(x-2)^2` => `(x-1)^2=(x-2)^2` => `(x-1)^2-(x-2)^2=0`

Use the formula for difference of squares: `m^2-n^2=(m-n)(m+n)`

`(x-1)^2-(x-2)^2=(x-1-x+2)(x-1+x-2)`

`(x-1)^2-(x-2)^2=(1)(2x-3)`

Put `(x-1)^2-(x-2)^2=0 ` => `2x-3=0`

Add 3 both sides => `2x=3` => `x=3/2`

Check the validity of this value.

`ln (3/2-1)^2= ln(1/4) `

`ln(3/2-2)^2=ln(-1/2)^2=ln(1/4)`

This value `x=3/2` checks both logarithms.

**ANSWER: The value of x that checks the identity is `x=3/2.` **