Solve for x given that lnx lnx 1 ln12 x 3 x 4 x 4 x 3

Solve for x, given that ln(x) + ln(x + 1) = ln(12). x = 3, x = -4 x = 4 x = 3, x = 4 x = 3 x = -3, x = -4 None of the above.

ln(x) + ln(x + 1) = ln(12)

Combining the lns on the left :
ln[x(x + 1)] = ln(12)

Now, cancel the ln\'s on both sides :
x(x + 1) = 12

x^2 + x = 12

x^2 + x - 12 = 0

(x + 4)(x - 3) = 0

x + 4 = 0 , x - 3 = 0

x = -4 and x = 3

Now, we need to check which of these works out...

We had original equation
ln(x) + ln(x+1) = ln(12)

When we plug in x = -4,
ln(x) becomes ln(-4) ---> undefined
So, x = -4 is extraneous, i.e invalid

No such problem happens with x = 3

So, we eliminate x = -4

And the only answer is :
x = 3 ---> ANSWER

Option D