For what value of x is: \log_10 (999+\sqrt{x^2-3x+3}) =3?

Logarithms Practice Test 1

Question : 38 of 40

Marks: +1, -0

For what value of x is:

\log_10 (999+√{x^2-3x+3}) =3?

Solution:

\log_10 (999+√{x^2-3x+3}) =3

⇒ 999+√{x^2-3x+3} =10^3

⇒√{x^2-3x+3}=1

⇒x^2-3x+3-1=0

⇒x^2-3x+2=0

⇒(x-1)(x-2)=0

Hence,

x=1,2

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