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dezoksy [38]
2 years ago
12

2logxbase5 3logybase3=8 and 6logxbase5 2logybase3=2 solve silmultaneously

Mathematics
1 answer:
Vinil7 [7]2 years ago
5 0
2\log_5{x}+3\log_3{y}=8 \ . \ . \ . \ (1)\\6\log_5{x}+2\log_3{y}=2 \ . \ . \ . \ (2)\\ \\ (1)\times3:6\log_5{x}+9\log_3{y}=24 \ . \ . \ . \ (3) \\ \\ (3)-(2):7\log_3{y}=22\\ \log_3y=\frac{22}{7}\\3^{\log_3y}=3^{\frac{22}{7}}=31.59\\y=31.59\\ \\ From \ (1), 2\log_5{x}+3\log_3{31.59}=8\\2\log_5{x}=8-9.429=-1.429\\ \log_5{x}= -\frac{1.429}{2} =-0.7143\\5^{\log_5{x}}=5^{-0.7143}\\x=0.3168

Therefore, x = 0.3168 and y = 31.59
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Answer:

f(x)=x^4-9x^2-50x-150

Step-by-step explanation:

Let f(x) be the polynomial function of minimum degree with real coefficients whose zeros are 5, -3, and -1 + 3i be f(x).

By the complex conjugate property of polynomials, -1-3i is also a root of this polynomial.

Therefore the polynomial in factored form is f(x)=(x-5)(x+3)(x-(-1+3i))(x-(-1+3i))

We expand to get:f(x)=(x^2-2x-15)(x^2+2x+10)

We expand further to get:\

f(x)=x^4-9x^2-50x-150

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3 years ago
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