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



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Posted on (Answer #1)

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




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)`



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) `


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

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

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