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Chapter 10: Problem 111

The solution of the equations \(\log x+\log x^{1 / 2}+\log x^{1 / 4}\) \(+\ldots=y\) and \(\frac{1+3+5+\ldots . .+(2 y-1)}{4+7+10+\ldots+(3 y+1)}\) \(=\frac{20}{7 \log x}\) is (A) \(x=10^{5}, 10^{-5 / 7}\) (B) \(y=10,-\frac{10}{7}\) (C) \(x=10,-\frac{10}{7}\) (D) \(y=10^{5}, 10^{-5 / 7}\)

Short Answer

Expert verified
Option A is correct: \(x=10^{5}, 10^{-5/7}\).

Step by step solution

01

Simplify the Logarithmic Expression

The first equation is given as \(\log x + \log x^{1/2} + \log x^{1/4} + \cdots = y\). This is an infinite series of logarithms. Recognize that \(\log x^{1/2} = \frac{1}{2}\log x\), \(\log x^{1/4} = \frac{1}{4}\log x\), and so on.We can express this as a sum of a series: \(\log x (1 + \frac{1}{2} + \frac{1}{4} + \cdots) = y\). This is a geometric series with the first term \(a = 1\) and a common ratio \(r = \frac{1}{2}\).
02

Sum the Infinite Geometric Series

The sum of an infinite geometric series \( 1 + r + r^2 + \cdots \) is given by \( S = \frac{a}{1 - r} \), where \( a \) is the first term and \( r \) is the common ratio. Here, \(a = 1\) and \(r = \frac{1}{2}\), so:\[ S = \frac{1}{1 - \frac{1}{2}} = \frac{1}{\frac{1}{2}} = 2 \]Therefore, substituting back into the equation, we have \(2 \log x = y\). Solving for \(x\), we get \( \log x = \frac{y}{2}\).
03

Simplify the Arithmetic Series

The second part of the problem involves the expression \( \frac{1 + 3 + 5 + \cdots +(2y-1)}{4 + 7 + 10 + \cdots +(3y+1)}\). Recognize that the numerator \(1 + 3 + 5 + \cdots +(2y-1)\) is the sum of the first \(y\) odd numbers, which is \(y^2\).
04

Simplify the Denominator Series

The denominator \(4 + 7 + 10 + \cdots +(3y+1)\) is an arithmetic series with the first term \(a = 4\), last term \(l = 3y + 1\), and common difference \(d = 3\). The number of terms is \(y\).The sum of an arithmetic series is \(S = \frac{y}{2}(4 + (3y + 1)) = \frac{y}{2}(3y + 5) \).
05

Solve the Fractional Equation

Plug in the simplified sums into the given equation:\[ \frac{y^2}{\frac{y}{2}(3y+5)} = \frac{20}{7 \log x} \]Simplify the left side: \( \frac{2y}{3y+5} \).Equating both sides, we have:\[ \frac{2y}{3y+5} = \frac{20}{7 \log x} \]
06

Substitute and Solve Simultaneous Equations

Replace \(\log x\) with \(\frac{y}{2}\) from Step 2:\[ \frac{2y}{3y+5} = \frac{20}{7 \times \frac{y}{2}} \]This simplifies to: \(\frac{2y}{3y+5} = \frac{40}{7y}\).Cross-multiplying gives: \(14y^2 = 40(3y+5)\), leading to the quadratic equation:\[ 14y^2 = 120y + 200 \]Solve for \(y\) by simplifying and factoring, or use the quadratic formula, and then solve for \(x\).
07

Solve for x and Verify Options

Continuing from the previous step, solve \( 14y^2 - 120y - 200 = 0 \) to find roots for \( y \). Check each possible case from the given options to determine which values satisfy both original equations.Calculate for each \(y\) potential values of \(x\) as \(x = 10^{y/2}\) based on \(\log x = \frac{y}{2}\), and verify amongst options. Choose the correct combination.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Logarithmic Series
A logarithmic series is a sequence where each term is the logarithm of a power of a number. In this exercise, we encounter an infinite logarithmic series that we have to simplify. The given series is
  • \( \log x + \log x^{1/2} + \log x^{1/4} + \cdots = y \).
To simplify such a series, recognize that each term can be expressed in multiples of \( \log x \):
  • \( \log x, \frac{1}{2}\log x, \frac{1}{4}\log x, \ldots \).
This sequence is actually a geometric series where the first term \( a = 1 \) and the common ratio \( r = \frac{1}{2} \). Geometric series are solved using the formula for the sum:
  • \[ S = \frac{a}{1 - r} \].
Substitute the values for our series:
  • \( S = \frac{1}{1 - \frac{1}{2}} = 2 \).
Thus, the sum of the series becomes \( 2 \log x = y \), leading us to solve for \( x \) by finding \( \log x = \frac{y}{2} \).
Geometric Series
The geometric series is a sequence where each term after the first is found by multiplying the previous one by a fixed number, called the "common ratio." In this exercise, our logarithmic series transforms into a geometric series with ratio \( \frac{1}{2} \). Here are some vital aspects of geometric series:
  • First term: this is the initial value of the series.
  • Common ratio \( r \): determines the pattern or growth between consecutive terms.
  • Infinite series sum: given by \( \frac{a}{1 - r} \) when \(|r|<1\).
In the context of our problem, the geometric series concept allows us to determine the sum of the infinite logarithmic series quickly and solve for critical values simplifying our solution.
Arithmetic Series
An arithmetic series is the sum of the terms of an arithmetic sequence, where each consecutive term is obtained by adding a constant difference to the previous one. In this task, we are concerned with two distinct arithmetic series, notably the denominator in our expression:
  • Numerator: sum of first \( y \) odd numbers, giving \( y^2 \).
  • Denominator: series \( 4, 7, 10, \ldots, (3y+1) \).
To find the sum of an arithmetic series, the formula is
  • \[ S = \frac{n}{2} (a + l) \], where \( n \) is the number of terms, \( a \) is the first term, and \( l \) is the last term.
For our problem, the series has first term \( a = 4 \), common difference \( d = 3 \), and last term \( l = 3y + 1 \). With these details, compute the sum for numerator and denominator which assists in solving the equation.
Quadratic Equations
The quadratic equation is a critical concept here as it emerges in the latter part of the exercise. Quadratic equations are polynomial equations of degree 2, typically expressed in the standard form:
  • \( ax^2 + bx + c = 0 \).
In this exercise, simplifying the expression leads us to the quadratic equation:
  • \( 14y^2 = 120y + 200 \)
Rewriting, we obtain:
  • \( 14y^2 - 120y - 200 = 0 \).
To solve this, use the quadratic formula:
  • \[ y = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \].
Here, \( a = 14 \), \( b = -120 \), and \( c = -200 \). Solving yields possible values for \( y \), which we then check against all original equations to find valid solutions for \( x \) making sure they fit into the conditions given in the options.

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Most popular questions from this chapter

The sum of first \(n\) terms of the series \(1^{2}+2.2^{2}+3^{2}+2.4^{2}+5^{2}+5.6^{2}+\ldots\) is \(\frac{n(n+1)^{2}}{2}\) when \(n\) is even. When \(n\) is odd, the sum is (A) \(\frac{n^{2}(n+1)}{2}\) (B) \(\frac{n(n+1)^{2}}{2}\) (C) \(\left[\frac{n(n+1)}{2}\right]^{2}\) (D) \(\frac{n(n+1)}{2}\)

If three positive numbers \(a, b, c\) are in H.P., then \(a^{n}+c^{n}\) \((\mathrm{A})>2 b^{n}\) \((\mathrm{B})=2 b^{n}\) \((\mathrm{C})<2 b^{n}\) \((\mathrm{D})>b^{n}\)

If \(H_{n}=1+\frac{1}{2}+\frac{1}{3}+\ldots .+\frac{1}{n}\), then the value of \(1+\frac{3}{2}+\frac{5}{3}+\ldots+\frac{2 n-1}{n}\) is (A) \(n-H_{n}\) (B) \(2 n-H_{n}\) (C) \((n-1)-H_{n}\) (D) \(n-2 H_{n}\)

If \(|a|<1\) and \(|b|<1\), then the sum of the series \(1+(1+a) b+\left(1+a+a^{2}\right) b^{2}+\left(1+a+a^{2}+a^{3}\right) b^{3}+\) \(\ldots \infty\) is equal to (A) \(\frac{1}{(1-b)(1-a b)}\) (B) \(\frac{1}{(1-a)(1-a b)}\) (C) \(\frac{1}{(1-a)(1-b)}\) (D) None of these

If the sum of first \(n\) terms of two A.P's are in the ratio \(3 n+8: 7 n+15\), then the ratio of their 12 th terms is (A) \(8: 7\) (B) \(7: 16\) (C) \(74: 169\) (D) \(13: 47\)
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