Question

Its urgent....

Solve :

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

Answer 1

$\boxed{\textbf{Answer : X = 5 }}$

$\textbf{\underline{#Step-by-step\: explanation.}}$

$log_{4}(x - 1) = log_{2}(x - 3) \\ \\ log_{4}(x - 1) = log_{4}( {x - 3}^{2} )$
▶As , Log4 base is same so we will equate the Argument both side.

$( {x - 3})^{2} = x - 1 \\ \\ {x}^{2} + 9 - 6x = x - 1 \\ \\ {x}^{2} - 7x + 10 = 0 \\ \\ (x - 2)(x - 5) = 0 \\ \\ x = 2 \\or \\ x = 5$

➡Here ,
⠀⠀⠀⠀⠀we get 2 values of x

•°• now Check in which value if x the log is defined.

x = 5 log is defined .

°•° X = 2 When we put then log(x - 3) is undefined)

So,

$\boxed{\textbf{Answer : X = 5 }}$

Answer 2

Answer:

1

Step-by-step explanation:

$Given:log_{4}(x - 1) = log_{2}(x - 3)$

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

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

$=>(x - 1) = (x - 3)^2$

$=>x - 1 = x^2 + 9 - 6x$

$=>x^2 - 7x + 10 = 0$

$=>x^2 - 5x - 2x + 10 = 0$

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

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

$=>x = 2,5$

When x = 5:

$=>log_{4}(5 - 1) = log_{2}(5 - 3)$

$=>log_{4}(4)=log_{2}(2)$

$=>1 = 1$

When x = 1:

$=>log_{4}(2 - 1) = log_{2}(2 - 3)$

Undefined.

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

Hope it helps!