log x + log(x-9)=1?

Answer 1

log 10 = 1

log 1 = 0

let x = 10

log x + log(x-9) = 1

log 10 + log 1 = 1

1 + 0 = 1

therefore x = 10

Answer 2

log x + log (x-9) = 1
log x(x-9) = 1

Put into its equivalent exponential form

10^1 = x(x - 9)
10 = x^2 - 9x
0 = x^2 - 9x - 10
0 = (x - 10)(x + 1)
x - 10 = 0
x = 10

and x + 1 = 0
x = -1, which can not be a solution because you can not take the log of a negative number

Answer 3

10^(log(x) + log(x - 9) = 10
10^log(x) * 10^log(x - 9) = 10

x *(x - 9) = 10
x^2 - 9x - 10 = 0
(x - 10)(x + 1) = 0

x = 10 or -1. But, x = -1 doesn't work, so x = 10.

Answer 4

You have to know:
a) If log(a)=x then 10^x=a (definition of logarithm)
b) logarithms exist only for positive numbers. (existance rule)
c) log (a*b)=log(a)+log(b) (product rule)
this means
x> and x-9>0, thus x>9.(existence rule)
Now log(x)+log(x-9)=1
or log (x(x-9))=1 (product rule)
10^1=x(x-9) (definition)
x²-9x=10 or
x² - 9x -10 = 0
(x-10)(x+1)=0 gives
x=10, the solution x=-1 is not admitted.(existence rule)

Answer 5

log x + log (x + 9) = log(x*(x + 9))

log(x*(x - 9)) = 1

log(x^2 - 9x) = 1 // 10^each side

x^2 - 9x = 10

x^2 - 9x - 10 = 0

x^2 - 2*4.5x + 4.5^2 - 4.5^2 - 10 = 0

(x - 4.5)^2 - 20.25 - 10 = 0

(x - 4.5)^2 - 30.25 = 0

(x - 4.5)^2 - 5.5^2 = 0

(x - 4.5 - 5.5)(x - 4.5 + 5.5) = 0

(x + 1)(x - 10) = 0

x = -1 is not a solution because log -1 is not real.

x=10?

log 10 + log(10 - 9) = 1 + log 1 = 1 + 0 = 1
Yes, x=10

Answer 6

Factor out log:
log(x(x-9))=1
x(x-9)=10
x^2-9x=10
x^2-9x-10=0
(x+1)(x-10)=0
x=10

Answer 7

log(x) + log (x-9) = log((x)*(x-9)
so the eqn above becomes ->
log(x*(x-9)) = 1
taking anilog,
x(x-9) = 10^1 or x(x-9) = 10
solve for x

Answer 8

tough, no way to solve it