Solve for x : log10 [log2 (log39)] = x -Maths 9th

Solve for x : log10 [log2 (log39)] = x -Maths 9th
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Evaluate x if log3 (3 + x) + log3 (8 – x) – log3 (9x – 8) = 2 – log39 -Maths 9th

Description : Evaluate x if log3 (3 + x) + log3 (8 – x) – log3 (9x – 8) = 2 – log39 -Maths 9th

Last Answer : (c) 4log3 (3 + x) + log3 (8 - x) - log3 (9x - 8) = 2 - log39 ⇒ log3 (3 + x) + log3 (8 - x) - log3 (9x - 8) + log39 = 2⇒ log3 \(\bigg[rac{(3+x)(8-x)(9)}{(9x-8)}\bigg]=2\)⇒ \(rac{9(24+8x-3x-x^2)}{(9x- ... = 9x - 8 ⇒ x2 + 4x - 32 = 0 ⇒ (x + 8) (x - 4) = 0 ⇒ x = - 8, 4. Taking the positive value x = 4.

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Description : Find the value of x, if log2 (5.2^x + 1), log4(2^(1–x) + 1) and 1 are in A.P. -Maths 9th

Last Answer : (b) 1 - log25 Given, log2 (5.2x + 1), log4 (21- x + 1), 1 are in A.P. ⇒ log2 (5.2x + 1) + 1 = 2 log4 (21 - x + 1) ⇒ log2 (5.2x + 1) + log22 = 2 log22 (21-x + 1)⇒ log2 (5.2x + 1).2 = 2 x \(rac12\) ... a=-rac{1}{2}\big)\)⇒ log 2x = log \(rac{2}{5}\)⇒ x log2 2 = log2 2 - log2 5 ⇒ \(x\) = 1 - log2 5.

If 5^(3x^2 log10 2) = 2^((x+1/2)log10 25), then the value of x is: -Maths 9th

Description : If 5^(3x^2 log10 2) = 2^((x+1/2)log10 25), then the value of x is: -Maths 9th

Last Answer : (d) \(-rac{1}{3}\)\(5^{{3x^2}log_{10}2}\) = 2\(\big(x+rac{1}{2}\big)\)log10 25⇒ \(5^{{3x^2}log_{10}2}\) = 2\(\big(rac{2x+1}{2}\big)\) x log10 5 = 2(2x+1)log10 5⇒ \(5^{{3x^2}log_{10}2}\) = 2(2x+1)log2 5. ... aloga x = x]⇒ 3x2 = 2x + 1 ⇒ 3x2 - 2x - 1 = 0 ⇒ (x - 1) (3x + 1) = 0⇒ x = 1, \(-rac{1}{3}\)

If log2 [log7(x^2 – x + 37)] = 1, then what could be the value of x ? -Maths 9th

Description : If log2 [log7(x^2 – x + 37)] = 1, then what could be the value of x ? -Maths 9th

Last Answer : (c) 4log2 [log7(x2 – x + 37)] = 1 ⇒ log7(x2 – x + 37) = 21 = 2 ⇒ x2 – x + 37 = 72 = 49 ⇒ x2 – x – 12 = 0 Now solve for x.

The number of meaningful solutions of log4(x – 1) = log2 (x – 3) is -Maths 9th

Description : The number of meaningful solutions of log4(x – 1) = log2 (x – 3) is -Maths 9th

Last Answer : (b) 1log4(x - 1) = log2(x - 3) ⇒ log22 (x − 1) = log2(x - 3)⇒ \(rac{1}{2}\) log2 (x-1) = log2 (x- 3) ⇒ log2 (x-1) = 2 log2 (x- 3)\(\big[\)Using logam (bn) = \(rac{n}{m} ... x = 2 or 5 Neglecting x = 2 as log2(x - 3) is defined when x > 2.⇒ There is only one meaningful solution of the given equation.

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Last Answer : (c) an irrational numberLet us assume log27 be a rational number. Then, log27 = \(rac{p}{q}\), where p,q ∈ I and q ≠ 0 ⇒ \(2^{rac{p}{q}}\) = 7 ⇒ 2p = 7q This is not true as 2 is even and 7 is odd.∴ Hence our assumption that log27 is a rational number is wrong. ∴ log27 is an irrational number.

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Last Answer : (d) \(a^{-rac{4}{3}}\)Since logaxa = \(rac{1}{log_aax}\) = \(rac{1}{log_aa+log_ax}\) = \(rac{1}{1+log_ax}\) andloga2x a = \(rac{1}{log_aa^2x}\) = \(rac{1}{log_aa^2+log_{ax}x}\) = \(rac{1}{2log_aa+log_{ax}x}\)= \( ... 2}}\)When t = \(-rac{4}{3}\), logax = \(-rac{4}{3}\) ⇒ \(x\) = \(a^{-rac{4}{3}}\)

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Last Answer : (b) \(\bigg(rac{10}{3},rac{20}{3}\bigg)\). (+ 10, 20) log100 |x+y| = \(rac{1}{2}\) ⇒ |x + y| = 100\(^{rac{1}{2}}\)⇒ |x + y| = 10 as (-10 is inadmissible) ...(i) log10y - log10| x | = log1004⇒ log10 ... x < 0, then x = 10.∴ If x = \(rac{10}{3}\), then y = \(rac{20}{3}\) and if x = 10, y = 20.

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A can solve 90% of the problems given in a book and B can solve 70%. What is the probability that at least one of them -Maths 9th

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Last Answer : Let E be the event that A solve the problem and F the event that B solves the problem.Then P(E) = \(rac{90}{100}\) = \(rac{9}{10}\), P(F) =\(rac{70}{100}\) = \(rac{7}{10}\), P(\(\bar{E}\) ... probability that at least one of them will solve a problem = 1 - \(rac{3}{100}\) = \(rac{97}{100}\) = 0.97.

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Derivation of Thiem's formula Q = 2 (s1 - s2)/2.3 log10 (r2/r1) is based on the assumption (A) The aquifer is homogeneous, isotropic and of infinite depth and area (B) The well is sunk through the full depth of the aquifer (C) The flow lines are radial and horizontal, and the flow is laminar (D) All the above

Description : Derivation of Thiem's formula Q = 2 (s1 - s2)/2.3 log10 (r2/r1) is based on the assumption  (A) The aquifer is homogeneous, isotropic and of infinite depth and area  (B) The well is sunk ... (C) The flow lines are radial and horizontal, and the flow is laminar  (D) All the above 

Last Answer : (D) All the above 

S1 and S2 are the draw downs in an observation well at times t1 and t2 after pumping. For discharge Q and coefficient of transmissibility T, the relationship, is (A) S2 - S1 = (2.3Q/ ) log10 (t2/t1) (B) S2 - S1 = (2.3Q/4 ) log10 (t2/t1) (C) S2 - S1 = (2.3Q/4 ) loge (t2/t1) (D) S2 - S1 = (2.3Q/4 ) loge (t1/t2

Description : S1 and S2 are the draw downs in an observation well at times t1 and t2 after pumping. For  discharge Q and coefficient of transmissibility T, the relationship, is  (A) S2 - S1 = (2.3Q/ ) log10 (t2/t1)  (B) S2 - S1 ... - S1 = (2.3Q/4 ) loge (t2/t1)  (D) S2 - S1 = (2.3Q/4 ) loge (t1/t2

Last Answer : (B) S2 - S1 = (2.3Q/4 ) log10 (t2/t1) 

he acentric factor of a materical, 'ω', is defined as ω = -log10(Prsat)Tr- 1 = 0.7, where, Prsat = reduced vapor pressure, Tr = reduced temperature. The value of acentric factor is always (A) > 2 (B) < 1 (C) > 1 (D) < 3

Description : he acentric factor of a materical, 'ω', is defined as ω = -log10(Prsat)Tr- 1 = 0.7, where, Prsat = reduced vapor pressure, Tr = reduced temperature. The value of acentric factor is always (A) > 2 (B) < 1 (C) > 1 (D) < 3

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the derivation of Thiem's formula, Q = (S1 - S2)/[2.3 log10( / )] the following assumption is not applicable (A) The aquifer is homogeneous and isotropic (B) Flow lines are radial and horizontal (C) The slope of the water surface is too small (D) The well has been sunk up to the surface of the unconfined aquifer

Description : the derivation of Thiem's formula, Q = (S1 - S2)/[2.3 log10( / )] the following assumption is not applicable (A) The aquifer is homogeneous and isotropic (B) Flow lines are radial and ... the water surface is too small (D) The well has been sunk up to the surface of the unconfined aquifer

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Description : p(x)=x3+3x2+3x+1, g(x) = x+2 -Maths 9th

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