Logarithm question? (multiple choice)

log (base 6) x + log (base 6) (x+2) = Log (base 6) 24

Find x

a) 11
b) 6
c) 4
d) -4
e) -6

log(base6) x + log(base6) (x+2) = log(base6) 24
log(base6) (x^2 + 2x) = log(base6)24
x^2 + 2x = 24
x^2 + 2x - 24 = 0
(x+6)(x-4) = 0
x = -6, x = 4

But you can't take the log of a negative number, so the answer is just 4.

c) 4

log(base 6) x + log(base 6) (x + 2) = log(base 6) 24

log(base 6) x(x + 2) = log(base 6) 24

x(x + 2) = 24

x^2 + 2x - 24 = 0

(x + 6)(x - 4) = 0

x = -6, 4

Note that x = -6 will not work since you cannot have a negative argument inside the log function. Therefore the answer is x = 4.

...sum of logs equals the log of the product !

log [base 6] x + log [base 6] (x + 2) = log [base 6] 24

log [base 6] (x)(x + 2) = log [base 6] 24

x^2 + 2x = 24

x^2 + 2x - 24 = 0

(x + 6)(x - 4) = 0

x = - 6 [extraneous]
or
x = 4

check

@ß

log(base 6)(x) + log(base 6)(x+2) = log(base 6)(24) --->

log(base 6)(x*(x+2)) = log(base 6)(24) ---> x(x+2) = 24 --->

x^2 + 2x - 24 = 0 ---> (x + 6)(x - 4) = 0, so either x = -6 or x = 4,

but since the domain of log is x > 0, the answer is x = 4, or c).