Logarithmic Function Problem?

Question

2 logx-log(x+1)=log4-log3

Can someone solve this?

Thanks.

Answer 1

We need two rules here:

(1) a.log(b) = log(b^a)
(2) log(a) - log(b) = log (a/b)

So 2 logx-log(x+1)=log4-log3

=> log(x²) - log(x+1)=log4-log3 (rule 1)

=> log[x²/(x+1)] = log(4/3) (rule 2 applied on both sides)

=> x²/(x+1) = 4/3

=> 3x² = 4(x+1)

=> 3x² - 4x -4 =0

=>(x-2)(3x+2)=0

=> x=2 or x=-2/3 BUT original equation was in logs and you can't take log(-2/3) - so this is not really a solution

=> x=2 is the only solution

Answer 2

The left hand side of the equation equals to
log(x^2 / (x-1)) and the right hand side equals to log(4/3), thus x^2/x-1 = 4/3

x²/(x+1) = 4/3

=> 3x² = 4(x+1)

=> 3x² - 4x -4 =0

=>(x-2)(3x+2)=0

=> x=2 or x=-2/3 you should ignore x=-2/3 because its log is not defined so x =2 is the only solution.

Answer 3

log x ² - log (x + 1) = log 4 - log 3
log [ x ² / (x + 1) ] = log (4 / 3)
x ² / (x + 1) = 4 / 3
3 x ² = 4 x + 4
3 x ² - 4 x - 4 = 0
( 3x + 2 ) ( x - 2 ) = 0
x = - 2 / 3 , x = 2
Accept + ve answer of x = 2

Answer 4

log[x^2/(x+1)] = log (4/3)
[x^2/(x+1)] = (4/3)
3x^2 = 4x + 1
3x^2 - 4x - 4 = 0
by factoring: (3x + 2)(x - 2)

3x + 2 = 0
[3x = -2]1/3
x = -(2/3)

x - 2 = 0
x = 2

answer is x = 2