log3(x+14)-log3(x+6)=log3x?

log₃(x + 14) - log₃(x + 6) = log₃x
log₃[(x + 14)/(x + 6)] = log₃x
(x + 14)/(x + 6) = x
x + 14 = x(x + 6)
x + 14 = x² + 6x
x² + 5x - 14 = 0
(x - 2)(x + 7) = 0
x = 2 or x = -7 (rejected, for log₃(-7) is undefined)

Answer: x = 2

log3 (x+14)-log3(x+6) = log3(x)
log3 ( (x+14)/(x+6) ) = log3(x)

(x+14)/(x+6) = x
multiply both sides by x+6
x+14 = x(x+6)
x+14 = x^2+6x
x^2+5x-14 = 0
(x+7)(x-2) = 0

x= -7 or x=2
x=2 is the answer because x=-7 cannot be plugged in log3(x+6)

x=2

Let's use subscripts to represent base 3.

so, log₃(x + 14) - log₃(x + 6) = log₃x

=> log₃[(x + 14)/(x + 6)] = log₃x

or, (x + 14)/(x + 6) = x

=> x + 14 = x(x + 6)

so, x² + 5x - 14 = 0

i.e. (x + 7)(x - 2) = 0

Therefore, x = -7 or x = 2

Now, for the log functions to exist, x + 14 > 0, x + 6 > 0 and x > 0

i.e. x > 0...must hold

so, x = 2 is the only valid solution.

:)>

Presuming that 3 is the base

log[3]((x + 14) / (x + 6)) = log[3](x)
(x + 14) / (x + 6) = x
x + 14 = x^2 + 6x
0 = x^2 + 5x - 14
x = (-5 +/- sqrt(25 + 56)) / 2
x = (-5 +/- 9) / 2
x = -14/2 , 4/2
x = -7 , 2

x = -7 doesn't work in the original problem

x = 2