Question

What is the true solution to the equation below? In e^in x + in e^in x^2 = 2 in 8

Answer 1

To solve the given equation, w need to review some rules:

`(1) \ ln \ e = 1 \\ (2) \ ln \ b^(a) = a*ln \ b \\ (3) \ ln \ a + ln \ b = ln \ (ab) \\ (4) \ ln \ a - ln \ b = ln \ (a)/(b) \\`

The given equation is :

`ln \ e^(ln \ x) + ln \ e^{ln \ x^(2)} = 2 \ ln \ 8`

`ln \ x * ln \ e + ln \ x^(2) * ln \ e = 2 \ ln \ 8` ⇒⇒⇒⇒ rule (2)

`ln \ x + ln \ x^(2) = ln \ 8^(2)` ⇒⇒⇒⇒ rule (1) and rule (2)


`ln \ ( x* x^(2) ) = ln \ 64` ⇒⇒⇒⇒ rule (3)
removing the nature logarithm from both sides


`x^(3) = 64 = 4^(3)`
∴ x = 4


So, the correct answer is option (2) ⇒⇒⇒⇒ x = 4