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)((15b - 15)/(-b^(2) + 1)) = 1`

`log_(2)(\frac{15(b) - 15(1)}{-1({b^(2)) - 1(-1)}}) = 1`

`log_(2)((15(b - 1))/(-1(b^(2) - 1))) = 1`

`log_(2)(-(15(b - 1))/(b^(2) - 1)) = 1`

`log_(2)(-(15(b - 1))/(b^(2) + b - b - 1)) = 1`

`log_(2)(-(15(b - 1))/(b(b) + b(1) - 1(b) - 1(1))) = 1`

`log_(2)(-(15(b - 1))/(b(b + 1) - 1(b + 1))) = 1`

`log_(2)(-(15(b - 1))/((b - 1)(b + 1))) = 1`

`log_(2)(-(15)/(b + 1)) = 1`

`2^(1) = -(15)/(b + 1)`

`2 = -(15)/(b + 1)`

`2(b + 1) = (b + 1)(-(15)/(b + 1))`

`2(b) + 2(1) = -15`

`2b + 2 = -15`

`2b = 13`

`(2b)/(2) = (-17)/(2)`

`b = -8.5`