Question

Log base 2 of (15b- 15) - log base 2 ( - b^2 + 1 ) = 1

Answer 1

$log_{2}(15b - 15) - log_{2}(-b^{2} + 1) = 1$
$log_{2}(\frac{15b - 15}{-b^{2} + 1}) = 1$
$log_{2}(\frac{15(b) - 15(1)}{-1({b^{2}) - 1(-1)}}) = 1$
$log_{2}(\frac{15(b - 1)}{-1(b^{2} - 1)}) = 1$
$log_{2}(-\frac{15(b - 1)}{b^{2} - 1}) = 1$
$log_{2}(-\frac{15(b - 1)}{b^{2} + b - b - 1}) = 1$
$log_{2}(-\frac{15(b - 1)}{b(b) + b(1) - 1(b) - 1(1)}) = 1$
$log_{2}(-\frac{15(b - 1)}{b(b + 1) - 1(b + 1)}) = 1$
$log_{2}(-\frac{15(b - 1)}{(b - 1)(b + 1)}) = 1$
$log_{2}(-\frac{15}{b + 1}) = 1$
$2^{1} = -\frac{15}{b + 1}$
$2 = -\frac{15}{b + 1}$
$2(b + 1) = (b + 1)(-\frac{15}{b + 1})$
$2(b) + 2(1) = -15$
$2b + 2 = -15$
$2b = 13$
$\frac{2b}{2} = \frac{-17}{2}$
$b = -8.5$