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Rom4ik [11]
3 years ago
5

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

Mathematics
2 answers:
wlad13 [49]3 years ago
8 0

Answer: OPTION B - X=4

Step-by-step explanation:

mariarad [96]3 years ago
5 0
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


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