Question

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

Answer 1

Answer: OPTION B - X=4

Step-by-step explanation:

Answer 2

To solve the given equation, w need to review some rules:
$(1) \ ln \ e = 1 \\ (2) \ ln \ b^{a} = a*ln \ b \\ (3) \ ln \ a + ln \ b = ln \ (ab) \\ (4) \ ln \ a - ln \ b = ln \ \frac{a}{b}$

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