Question

15. The number of solution (s) of the equation
log2 (x² - 1) = log1/2(x-1)= a real number, is
(a)0
(b) 1
(c)2
(d) 3
​

Answer 1

$log_{2}(x {}^{2} - 1 ) = log_{1 \div 2}(x - 1) \\ \\ log_{2}(x {}^{2} - 1) = log_{2 {}^{ - 1} }(x - 1) \\ \\ log_{2}(x {}^{2} - 1) = - log_{2}(x - 1) \\ becoz \: log_{y {}^{n} }(u) = (1 \div n) log_{y}(u) \\ \\ log_{2}(x {}^{2} - 1) = log_{2}(x - 1) {}^{ - 1} \\ becoz \: \: log(t {}^{n} ) = n log(t) \\ \\ log_{2}(x {}^{2} - 1 ) - log_{2}(x - 1) {}^{ - 1} = 0 \\ \\ log_{2}(x + 1) = 0 \\ becoz \: \: log(mn) = log(m) + log(n) \\ \\ (x + 1) = 2 {}^{0} \\ becoz \: \: log_{ \alpha }( \beta ) = y \\ \beta = \alpha {}^{y} \\ \\ x + 1 = 1 \\ \\ x = 0$

Answer 2

Answer:

1

Step-by-step explanation: