logarithms?

Question

logx + log(x-4) = log5

solve for x

i've started the problem but got stuck

log (x(x-5)) = log 5
log (x^2 - 5x) = log 5
(x^2 - 5x) = 5

Answer 1

logx + log(x-4) = log 5
log(x * [x - 4]) = log 5
log (x² - 4x) = log 5
x² - 4x = 5
x² - 4x - 5 = 0
Now get the linear factors.
(x - 5)(x + 1) = 0
x = 5 or x = -1
Check both values.
x = 5 is the Answer.

Answer 2

logx + log(x-4) = log5
log (x(x-4)) = log 5
log (x^2 - 4x) = log 5
(x^2 - 4x) = 5
x^2 - 4x - 5 = 0
x^2 -5x + x - 5 = 0
x(x-5)+1(x-5)=0
(x-5)(x+1)=0
x=5
x=-1

Disregard x=-1 ---- log(-1) is not a real number
x=5

Your mistake was converting the (x-4) to (x-5) in the first step.

Answer 3

it should be x^2 - 4x = 5

subtract 5 for both sides
x^2 0- 4x - 5 = 0

factor
(x - 5) (x + 1) = 0

x = 5 or -1

there can not be a negative for log. The only solution to this problem is 5

Answer 4

is 4 not 5


log (x(x-4)) = log 5
log (x^2 - 4x) = log 5
(x^2 - 4x) = 5
x^2 - 4x - 5 = 0
(x - 5)(x +1) = 0
x = 5 , -1

Answer 5

log x + log(x-4) = log5

log (x(x-4)) = log 5

log (x² -4x) = log 5

x² - 4x = 5

x² - 4x -5 = 0

Use the quadratic formula.
x = [4±√16-4(-5)]/2

x = (4±√36)/2

x = 2±3

x = 5 or -1

but log(-1) is not a real number, so it is rejected.
in fact, log(-1) = iπ

Answer 6

log[ (x).(x - 4) ] = log 5
(x).(x - 4) = 5
x² - 4x - 5 = 0
(x - 5).(x + 1) = 0
x = 5 , x = - 1