Question

4. Solve the following logarithmic equation and find the extraneo root: a. logs (x 6) + logs(x + 2) 1 b. Inz Inćz +6)- Incr -4)

Answer 1

Given,

a. log₅(x+6) +  log₅​(x+2) = 1

We know that log _aa = 1. So, in this case, log₅ 5 = 1

and log a + log b = log ab

Using the above formulas, we get

log₅(x+6)(x+2) = log₅ 5

Since both logarithms are equal, equating the values, we get

(x+6)(x+2) = 5

x²+8x+12 = 5

x²+8x+7 = 0

(x+7)(x+1) = 0

x = -7 or -1

Since the negative logarithm is not possible, x can't take the value as -7. So,

x = -1 is the answer.

b) ln x = ln (x+6) - ln (x-4)

So, here

x(x-4) = x+6

x²-4x-x-6 = 0

x²-5x-6 = 0

x²-6x+x-6 = 0

(x-6)(x+1) = 0

x = 6 or -1

Since -1 is not possible as it is not defined for negative values,

x = 6 is the answer.