Solve for X with Logs?

log(x - 3) + log(x) = 1
log[x(x - 3)] = 1
x(x - 3) = 10^1
x^2 - 3x - 10 = 0
x^2 + 2x - 5x - 10 = 0
(x^2 + 2x) - (5x + 10) = 0
x(x + 2) - 5(x + 2) = 0
(x + 2)(x - 5) = 0

x + 2 = 0
x = -2

x - 5 = 0
x = 5

But...

log(-2) is invalid.

∴ x = 5

Then (x^2-3x)=10
x^2-3x-10=0
(x-5)(x+2)=0
x=5
God bless you.

Remember that: log(a) + log(b) = log(ab)
log(x(x - 3)) = 1
log(x^2 - 3x) = 1

Also, recall the relationship: If logₐx = y, then a^y = x.
10^1 = x^2 - 3x
10 = x^2 - 3x
0 = x^2 - 3x - 10
0 = (x - 5)(x + 2)

Hence, we have 5 and -2. However, substituting -2 in place of x in the logarithmic equation will force us to take the logarithm of a negative number, which is invalid. Hence, x can only be 5.

Hope this helps!

log(x-3)+log(x)=1
= log(x^2-3x)=1
=10^1=x^2-3x
=10=x^2-3x

x=5