math question?

Question

how do I do this?

solve the logarithmic equation:
log x + log(x-9)=1

Answer 1

log[x(x-9)] = 1

10^1 = x(x-9)

10 = x^2 - 9x

0 = x^2 - 9x - 10

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

x= 10 or x = -1

-1 is not in the domain of this equation, so x = 10

Answer 2

log(x) + log(x-9)=1 Original equation
log(x(x-9))=1 Addition property of logs [log(a)+log(b)=log(a*b)]
x(x-9)=10 Raise 10 to the power of both sides. 10^log(x)=x
x^2-9x=10 Distribute the x
x^2-9x-10=0 Rearrange into ax^2+bx+c form
x=10 x=-1 Use the quadratic formula
x=-1 isn't in the domain of the log function, so x=10 is the only answer.

Answer 3

log x + log(x-9)=1

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

Now raise each side to the power of 10

10^(log(x * (x-9))) = 10^1

x*(x-9) = 10

x^2 - 9x - 10 = 0

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

Hence x =10 or -1

But you cannot take the log of a negative number, hence x = 10

Answer 4

log x = log (base 10) x RIGHT?

Good, now you need to raise 10 to the log (base 10) x and do the same on the other side.

10^log x => x do you see that?

If it were ln x that is the same as ln (base e) x

To change that we would make raise {e} the base to the {log x}
Like this: e^ln x = x
................................

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

10 ^ (log x) + 10 ^ (log (x-9)) = 10 ^1

This becomes

x + (x-9) = 10^1

2x -9 = 10

2x = 19

x = 19/2

I hope you learned something