log(2x-1)+log(x+4)=1?

1 ⤊

log(2x-1)+log(x+4)=1?

Perform this operation

2007-12-04 19:02:21 · 3 answers · asked by monkeybrat12006 1 in Science & Mathematics ➔ Mathematics

3 answers

Let logs be to base 10:-
log (2x - 1)(x + 4) = 1
(2x - 1)(x + 4) = 10
2x² + 7x - 4 = 10
2x² + 7x - 14 = 0
x = [- 7 ± √(49 + 112) ] / 4
x = [- 7 ± √(161) ] / 4
x = 1.42 , x = - 4.92

2007-12-04 19:54:57 · answer #1 · answered by Como 7 · 1⤊ 0⤋

Exponentiate both sides, assuming log base 10:
(2x-1)(x+4) = 10. Expand and simplify:
2x^2 +7x - 14 = 0. Solve, using the quadratic formula; the roots are irrational.

2007-12-04 19:09:28 · answer #2 · answered by Anonymous · 0⤊ 0⤋

log[(2x-1)(x+4)]=1
considering log to the base 10,
(2x-1)(x+4)=10
2x^2+7x-4=10
2x^2+7x-14=0
now use [-b+-sqrt(b^2-4ac)]/2a

2007-12-04 19:16:53 · answer #3 · answered by naky 1 · 0⤊ 0⤋