HOW do you solve log6 (X+1) = 1 - log6 X?

Question

log₆ (X+1) = 1 - log₆ X

Answer: 2

Answer 1

Let’s try moving the logs together, by adding log₆X to each side.

log₆(X+1) + log₆(X) = 1

Now combine the left side, as the sum of the logs is the log of the product:

log₆(X(X+1)) = 1

now raise 6 to these powers:

6^log6(X(X+1)) = 6¹

(X(X+1)) = 6

X² + X = 6

Move to left hand side

X² + X - 6 = 0

(X + 3)(X - 2) = 0

So X = 2 or -3

but logarithm is not defined for negative arguments, so X = 2 is the answer.

Answer 2

Recall the rules:

1) log A + log B = log(A+B)

2) log A - log B = log(A/B)

3) if log M = log N then M=N

4) if logB(1)=0, so if logB(k)=0 then k=1

5) if logB(B) = 1 , so if logB(k)=1, the k=B

log(x+1) = 1 - log(x) <--- given

log(x+1) + log(x) = 1 <--- adds log x to both sides

log[(x+1)x] = 1 <--- LOG (a) + log(b) = log(a+b) as shown above in RUle #1

(x+1)x = 6 <--- if log(k)=1 then k=6, as shown above in Rule #5

x^2 + x - 6 = 0 <--- distributes the x and moves 6 to left side

( x + 3 )( x - 2 ) = 0 <--- factors

x+3=0 ---> x = -3

x - 2 = 0 ---> x = 2

CHECK:

x=-3 is disqualified because x+1 = -3 +1 = -2 are x=-2 are

NOT in the domain of the log function, per the given equation.

cannot take log of negative number or zero,

FOr x=2

log(2+1) = 1 - log 2

log(3) = 1 - log 2

log(3) = log(6) - log2

log(3) = log(6/2)

log(3) = log(3) <--- yes it checks

or

log(2+1) = 1 - log(2)

log(3) = 1 - log(2)

log(3) + log(2) = 1

log(3*2) = 1

log(6) = 1

1 = 1 <--- yes it checs