+129270 log5x+log(x-1)= 2 Note....... 2 = log 100
log (5x * (x -1) ) = log 100 and we can now just ignore the logs and solve directly
5x ( x - 1) = 100 simplify
5x^2 - 5x = 100
5x^2 - 5x - 100 = 0 divide through by 5
x^2 - x - 20 = 0 factor
(x - 5) ( x + 4) = 0 and setting both factors to 0 we have that x = 5 or x = -4
We must reject -4 because it makes the original logs negative......and that's undefined.....thus......
x = 5 is the solution
+129270 log5x+log(x-1)= 2 Note....... 2 = log 100
log (5x * (x -1) ) = log 100 and we can now just ignore the logs and solve directly
5x ( x - 1) = 100 simplify
5x^2 - 5x = 100
5x^2 - 5x - 100 = 0 divide through by 5
x^2 - x - 20 = 0 factor
(x - 5) ( x + 4) = 0 and setting both factors to 0 we have that x = 5 or x = -4
We must reject -4 because it makes the original logs negative......and that's undefined.....thus......
x = 5 is the solution
