Please tell me what is log base 3(x^2 −3x−31)=2 and 2 log x=log 2+log(4x − 6)
Given
logbasea(a) =1
2logbasea(a) = logbasea(a^2)
so
2logbase3(3) = logbase3(3^2)
log base 3(x^2 −3x−31)=logbase3(3^2)
this means
x^2 −3x−31= 3^2
x^2 −3x−40= 0
Factoring
(x - 8)(x + 5) = 0
x = -8, x = -5
2 log x=log 2+log(4x − 6).................(log a + log b = log ab)
log x^2 = log 2(4x -6)
x^2 = 2(4x -6)
x^2 -8x + 12 = 0
(x - 6)(x - 2) = 0
x = 6 , x = 2
log(base 2) (2x-4) = 2 + log(base 2) (x^2-6) => log(base 2) (2x-4) - log(base 2) (x^2-6) = 2 => log(base 2) [(2x - 4) / (x^2 - 6) ] = 2 => (2x - 4) /(x^2 - 6) = 2^2 = 4 => 2x - 4 = 4x^2 - 24 => divide by 2 x - 2 = 2x^2 - 12 => 2x^2 - x - 10 = 0 => 2x^2 - 5x + 4x - 10 = 0 => x(2x - 5) + 2(2x - 5) = 0 => (2x - 5)(x + 2) = 0 since log of negative numbers is not defined 2x = 5 x = 5/2
Copyright © 2024 1QUIZZ.COM - All rights reserved.
Answers & Comments
Verified answer
Given
logbasea(a) =1
2logbasea(a) = logbasea(a^2)
so
2logbase3(3) = logbase3(3^2)
log base 3(x^2 −3x−31)=logbase3(3^2)
this means
x^2 −3x−31= 3^2
x^2 −3x−40= 0
Factoring
(x - 8)(x + 5) = 0
x = -8, x = -5
2 log x=log 2+log(4x − 6).................(log a + log b = log ab)
log x^2 = log 2(4x -6)
so
x^2 = 2(4x -6)
x^2 -8x + 12 = 0
(x - 6)(x - 2) = 0
so
x = 6 , x = 2
log(base 2) (2x-4) = 2 + log(base 2) (x^2-6) => log(base 2) (2x-4) - log(base 2) (x^2-6) = 2 => log(base 2) [(2x - 4) / (x^2 - 6) ] = 2 => (2x - 4) /(x^2 - 6) = 2^2 = 4 => 2x - 4 = 4x^2 - 24 => divide by 2 x - 2 = 2x^2 - 12 => 2x^2 - x - 10 = 0 => 2x^2 - 5x + 4x - 10 = 0 => x(2x - 5) + 2(2x - 5) = 0 => (2x - 5)(x + 2) = 0 since log of negative numbers is not defined 2x = 5 x = 5/2