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)`
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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