Question

100 points.

If log base 2(x-1)+log base2(x+1) = 3.
Then find x..

Anybody plzz answer......​

Answer 1

$\huge \boxed{ \underline{ \underline{ \bf{Answer}}}}$

GIVEN :

$\sf{log_{2}(x-1) + log_{2}(x+1) = 3}$

TO FIND :

Value of 'x'.

SOLUTION :

$\implies \sf{log_{2}(x-1) + log_{2}(x+1) = 3}$

By using Identity log(a)+log(b)= log(ab)

$\implies \sf{log_{2}(x-1)(x+1) = 3}$

$\implies \sf{log_{2}( {x}^{2} - {1}^{2} ) = 3}$

$\implies \sf{log_{2}( {x}^{2} - {1} ) = 3}$

$\implies \sf{ \cancel{log}({x}^{2} - 1) = {2}^{3} }$

$\implies \sf{{x}^{2} - 1 = 8}$

$\implies \sf{{x}^{2} = 9}$

$\sf{ \implies \: x = 3 \: \: \: or \: \: \: x = - 3(not \: possible)}$

$\boxed{ \sf{Hence, \: Value \: of \: x = 3 }}$

Answer 2

Step-by-step explanation:

$log_{2}(x - 1) + log_{2}(x + 1) = 3 \\ \\ log_{2}(x - 1) (x + 1) = 3 \\ \\ log_{2}( {x}^{2} - 1 ) = 3 \\ \\ now \\ \\ {2}^{3} = {x}^{2} - 1 \\ \\ 8 = {x}^{2} - 1 \\ \\ {x}^{2} - 1 - 8 = 0 \\ \\ {x}^{2} - 9 = 0 \\ \\ {x}^{2} - {3}^{2} = 0 \\ \\ (x + 3)(x - 3) = 0 \\ \\ \\ therefore \\ \\ x = 3 \: \: \: or \: x = - 3$

Since, negative value is not considerd

Therefore, x = 3

Identities used :

>> log a+ log = log ab

>> a²-b² = (a+b)(a-b)

>> log base a (N) = x => a^x = N

Hope it helps..