Solve the logarithmic equation.
log2(x-1) - log2(5x + 1) = -3
log2[(x-1)(5x + 1)] = -3
(x-1)(5x + 1) = 2^-3
5x^2 - 5x - 1 = 1/8
5x^2 + 4x - 7/8 = 0

This class is killing me.

Question

Solve the logarithmic equation.
log2(x-1) - log2(5x + 1) = -3
log2[(x-1)(5x + 1)] = -3
(x-1)(5x + 1) = 2^-3
5x^2 - 5x - 1 = 1/8
5x^2 + 4x - 7/8 = 0

This class is killing me.

anonymous · Answer

I was trying to follow along with my book. But of course the only give easy examples.

anonymous · Answer

is it the log to base 2?

anonymous · Answer

log2+log(x-1)-log2-log(5x+1)=-3
or,  log(x-1)-log(5x+1)=-3
or,   log((x-1)/(5x+1))=-3
or,   (x-1)/(5x+1)=2pow(-3)=1/8
now try to find the value of x

anonymous · Answer

|dw:1320937198332:dw|yes.