Question

The number of solutions of log4 ( x - 1 ) = log2 ( x - 3 ) is

Answer 1

$log_{4} (x-1)=log_{2} (x-3) \\ \frac{1}{2} log_{2} (x-1)=log_{2} (x-3) \\ log_{2} (x-1)=2log_{2} (x-3) \\ log_{2} (x-1)=log_{2} (x-3)^2 \\x-1=(x-3)^2 \\ x^{2} -6x+9-x+1=0 \\ x^{2} -7x+10=0 \\ (x-5)(x-2)=0 \\ x=5 (x \neq 2)$
as log (-1) is not defined..

Answer 2

Answer:

1

Step-by-step explanation:

Given:log4(x−1)=log2(x−3)

= > log_{4}(x - 1) = 2log_{4}(x - 3)=>log4(x−1)=2log4(x−3)

= > log_{4}(x - 1) = log_{4}(x - 3)^2=>log4(x−1)=log4(x−3)2

= > (x - 1) = (x - 3)^2=>(x−1)=(x−3)2

= > x - 1 = x^2 + 9 - 6x=>x−1=x2+9−6x

= > x^2 - 7x + 10 = 0=>x2−7x+10=0

= > x^2 - 5x - 2x + 10 = 0=>x2−5x−2x+10=0

= > x(x - 5) - 2(x - 5) = 0=>x(x−5)−2(x−5)=0

= > (x - 2)(x - 5) = 0=>(x−2)(x−5)=0

= > x = 2,5=>x=2,5

When x = 5:

= > log_{4}(5 - 1) = log_{2}(5 - 3)=>log4(5−1)=log2(5−3)

= > log_{4}(4)=log_{2}(2)=>log4(4)=log2(2)

= > 1 = 1=>1=1

When x = 1:

= > log_{4}(2 - 1) = log_{2}(2 - 3)=>log4(2−1)=log2(2−3)

Undefined.

Therefore, the value of x = 5. Hence, It has 1 solution.