Solve for x and y...?

fardeen_gen · #1 $Old$ November 6th, 2008, 09:38 AM

log(base 10) x + log(base 10) x^1/2 + log(base 10) x^1/4 + ... to ∞ = y
(1 + 3 + 5 + 7 + ... +(2y - 1))/(4 + 7 + 10 + ... + (3y + 1)) = 20/(7.log(base 10) x)

Plato · #2 $Old$ November 6th, 2008, 10:05 AM

fardeen_gen · #3 $Old$ November 6th, 2008, 10:29 AM

Sorry, couldn't understand that

fardeen_gen · #4 $Old$ November 10th, 2008, 05:34 AM

Is this correct?

log(base 10) x + log(base 10) x^1/2 + log(base 10) x^1/4 + ... to ∞ = y... 1

(1 + 3 + 5 + 7 + ... +(2y - 1))/(4 + 7 + 10 + ... + (3y + 1)) = 20/(7.log(base 10) x) ...2

as there are y terms in the 2nd expression y is integer

let us take log(base 10) x = t .. 3

then log (base 10) x^(1/k) = 1/k log (base 10)

so 1st equation becomes

t(1+1/2+ 1/4 ....) = y

or 2t= y .... 4

now for the 2nd equation
numerator(this is AP)

1+3 + 5 + .. + (2y-1) = y^2 ... 4

denominator(another AP)

4+7 + ... (3y+1) = 3y(y+1)/2 + y ..5

so

LHS = y^2/(3y(y+1)/2 + y) = 2y/(3y+5) = 20/7t
y = 2t gives

4t/(6t + 5) = 20/7t

or t/(6t+5) = 5/7t

or 7t^2 = 30 t + 25
or 7t^2 - 30 t - 25 = 0

(7t - 35 t + 5t - 25 = 0

(7t-5)(t-5) = 0

so t = 5 and y = 10 as t = 5/7 shall make y non integer

so x = 10^5

so ans x = 10^5 and y = 10

Soroban · #5 $Old$ November 10th, 2008, 06:42 AM

Hello, fardeen_gen!

Quote:

$\log(x) + \log\left(x^{\frac{1}{2}}\right) + \log\left(x^{\frac{1}{4}}\right) + \hdots \;=\;y$ .[1]

$\frac{1 + 3 + 5 + 7 + \hdots +(2y - 1)}{4 + 7 + 10 + \hdots + (3y + 1)} \;= \;\frac{20}{7\log(x)}$ .[2]

Equation [1]: . $\log\left(x\cdot x^{\frac{1}{2}}\cdot x^{\frac{1}{4}}\cdot x^{\frac{1}{8}}\cdots\right) \:=\:y\quad\Rightarrow\quad \log(x^2) \:=\:y$

. . . $2\log(x) \:=\:y \quad\Rightarrow\quad \log(x) \:=\:\frac{y}{2}$ .[3]

In Equaton [2], the numerator is an arithmetic series
with first term $a = 1$ , common difference $d = 2$ , and $y$ terms.
. . Its sum is: . $y^2$

The denominator is an arithmetic series
with first term $a = 4$ , common difference $d = 3$ , and $y$ terms.
. . Its sum is: . $\frac{y}{2}(3y+5)$

Hence, [2] becomes: . $\frac{y^2}{\frac{y}{2}(3y+5)} \;=\;\frac{20}{7\log(x)} \quad\Rightarrow\quad 7y\log(x) \:=\:30y + 50$ .[4]

Substitute [3] into [4]: . $7y\cdot\frac{y}{2} \:=\:30y + 50 \quad\Rightarrow\quad 7y^2 - 60y - 100 \:=\:0$

Factor: . $(y - 10)(7y + 10) \:=\:0 \quad\Rightarrow\quad y \:=\:10,\;\text{-}\,\tfrac{10}{7}$

Substitute into [3]: . $\begin{array}{ccc}\log(x) \:=\:\frac{10}{2} & \Rightarrow & x \:=\:10^5 \\ \log(x) \:=\:\text{-}\frac{5}{7} & \Rightarrow & x \:=\:10\,^{\text{-}\frac{5}{7}} \end{array}$

Solutions: . $\left(10^5,\:10\right),\;\;\left(10^{-\frac{5}{7}},\:\text{-}\,\tfrac{10}{7}\right)$