Q7. Find the sum of all integers satisfying the inequalities...
Answers
EXPLANATION.
Sum of all integers satisfying the inequalities.
⇒ ㏒₅(x - 3) + 1/2㏒₅(3) < 1/2㏒₅(2x² - 6x + 7) and ㏒₃(x) + ㏒_(√3)(x) + ㏒_(1/3) (x) < 6.
As we know that,
⇒ (x - 3) > 0.
⇒ x > 3. - - - - - (1).
Equation (1).
⇒ ㏒₅(x - 3) + 1/2㏒₅(3) < 1/2㏒₅(2x² - 6x + 7).
⇒ ㏒₅(x - 3) < 1/2㏒₅(2x² - 6x + 7) - 1/2㏒₅(3).
As we know that,
Formula of :
⇒ ㏒ₐM - ㏒ₐN = ㏒ₐM/N.
Using this formula in equation, we get.
⇒ ㏒₅(x - 3) < 1/2㏒₅[2x² - 6x + 7/(3)].
⇒ 2㏒₅(x - 3) < ㏒₅[2x² - 6x + 7/3].
As we know that,
Formula of :
⇒ x㏒ₐN = ㏒ₐNˣ. (x any real number).
Using this formula in equation, we get.
⇒ ㏒₅(x - 3)² < ㏒₅[2x² - 6x + 7/3].
⇒ 3(x - 3)² < [2x² - 6x + 7].
As we know that,
Formula of :
⇒ (x - y)² = x² + y² - 2xy.
⇒ 3(x² + 9 - 6x) < [2x² - 6x + 7].
⇒ 3x² + 27 - 18x < 2x² - 6x + 7.
⇒ 3x² + 27 - 18x - 2x² + 6x - 7 < 0.
⇒ x² - 12x + 20 < 0.
Factorizes the equation into middle term splits, we get.
⇒ x² - 10x - 2x + 20 < 0.
⇒ x(x - 10) - 2(x - 10) < 0.
⇒ (x - 2)(x - 10) < 0.
As we know that,
Find the zeroes and the point on wavy curve method, we get.
⇒ x - 2 = 0.
⇒ x = 2.
⇒ x - 10 = 0.
⇒ x = 10.
We get,
⇒ x ∈ (2,10). - - - - - (2).
Equation (2).
⇒ ㏒₃(x) + ㏒_(√3)(x) + ㏒_(1/3)(x) < 6.
As we know that,
We can write equation as,
⇒ ㏒₃(x) + ㏒_(3)^1/2(x) + ㏒₃₍₋₁₎(x) < 6.
⇒ ㏒₃(x) + 2㏒₃(x) - ㏒₃(x) < 6.
⇒ 2㏒₃(x) < 6.
⇒ ㏒₃(x) < 3.
As we know that,
Formula of :
⇒ x㏒ₐN = ㏒ₐNˣ. (x any real number).
⇒ x < (3)³.
⇒ x < 27. - - - - - (3).
From equation (1) & (2) & (3), we get.
(1) = x > 3. - - - - - (1).
(2) = x ∈ (2,10). - - - - - (2).
(3) = x < 27. - - - - - (3).
⇒ x ∈ (3,4,5,6,7,8,9).
As we all know that,
From equation (1) x > 3 not equal to 3 that's why 3 is not considered.
⇒ x ∈ (4,5,6,7,8,9).
Number of integers = 6.
Sum of integers = 4 + 5 + 6 + 7 + 8 + 9 = 39.
MORE INFORMATION.
Logarithmic inequality.
Let a is real number such that,
(1) = For a > 1 the inequality ㏒ₐx > ㏒ₐy & x > y are equivalent.
(2) = If a > 1 then ㏒ₐx < n ⇒ 0 < x < aⁿ.
(3) = If a > 1 then ㏒ₐx > n ⇒ x > aⁿ.
(4) = For 0 < a < 1 the inequality 0 < x < y & ㏒ₐx > ㏒ₐy are equivalent.
(5) = If 0 < a < 1 then ㏒ₐx < n ⇒ x > aⁿ.
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