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Question Number 144563 by mathdanisur last updated on 26/Jun/21
log_4 (x) = log(y) = log_(25) (x+y) Find (x/y) = ?
$$\boldsymbol{{log}}_{\mathrm{4}} \left(\boldsymbol{{x}}\right)\:=\:\boldsymbol{{log}}\left(\boldsymbol{{y}}\right)\:=\:\boldsymbol{{log}}_{\mathrm{25}} \left(\boldsymbol{{x}}+\boldsymbol{{y}}\right) \\ $$$${Find}\:\:\frac{{x}}{{y}}\:=\:? \\ $$
Answered by imjagoll last updated on 26/Jun/21
let log _4 (x)=log _(10) (y)=log _(25) (x+y)=k then { ((x=4^k )),((y=10^k )),((x+y=25^k )) :} ⇔ 4^k + 10^k = 25^k ⇒1+((5/2))^k =((5/2))^(2k) ⇒t^2 −t−1 =0 ; where t=((5/2))^k ⇒t=((1+(√5))/2) ⇒we find (x/y)= (4^k /(10^k ))=((2/5))^k =(2/( (√5)+1)) (x/y)= ((2((√5)−1))/4)= (((√5)−1)/2)
$$\:\mathrm{let}\:\mathrm{log}\:_{\mathrm{4}} \left(\mathrm{x}\right)=\mathrm{log}\:_{\mathrm{10}} \left(\mathrm{y}\right)=\mathrm{log}\:_{\mathrm{25}} \left(\mathrm{x}+\mathrm{y}\right)=\mathrm{k} \\ $$$$\mathrm{then}\:\begin{cases}{\mathrm{x}=\mathrm{4}^{\mathrm{k}} }\\{\mathrm{y}=\mathrm{10}^{\mathrm{k}} }\\{\mathrm{x}+\mathrm{y}=\mathrm{25}^{\mathrm{k}} }\end{cases} \\ $$$$\Leftrightarrow\:\mathrm{4}^{\mathrm{k}} \:+\:\mathrm{10}^{\mathrm{k}} \:=\:\mathrm{25}^{\mathrm{k}} \\ $$$$\Rightarrow\mathrm{1}+\left(\frac{\mathrm{5}}{\mathrm{2}}\right)^{\mathrm{k}} =\left(\frac{\mathrm{5}}{\mathrm{2}}\right)^{\mathrm{2k}} \\ $$$$\Rightarrow\mathrm{t}^{\mathrm{2}} −\mathrm{t}−\mathrm{1}\:=\mathrm{0}\:;\:\mathrm{where}\:\mathrm{t}=\left(\frac{\mathrm{5}}{\mathrm{2}}\right)^{\mathrm{k}} \\ $$$$\Rightarrow\mathrm{t}=\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{we}\:\mathrm{find}\:\frac{\mathrm{x}}{\mathrm{y}}=\:\frac{\mathrm{4}^{\mathrm{k}} }{\mathrm{10}^{\mathrm{k}} }=\left(\frac{\mathrm{2}}{\mathrm{5}}\right)^{\mathrm{k}} =\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}+\mathrm{1}} \\ $$$$\frac{\mathrm{x}}{\mathrm{y}}=\:\frac{\mathrm{2}\left(\sqrt{\mathrm{5}}−\mathrm{1}\right)}{\mathrm{4}}=\:\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}} \\ $$
Commented by mathdanisur last updated on 26/Jun/21
alot cool thanks Sir, answer (1/2)((√5)+1)
$${alot}\:{cool}\:{thanks}\:{Sir},\:{answer}\:\frac{\mathrm{1}}{\mathrm{2}}\left(\sqrt{\mathrm{5}}+\mathrm{1}\right) \\ $$
Commented by imjagoll last updated on 26/Jun/21
wrong. if (y/x)= (((√5)+1)/2) true
$$\mathrm{wrong}.\:\mathrm{if}\:\frac{\mathrm{y}}{\mathrm{x}}=\:\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{2}}\:\mathrm{true} \\ $$
Commented by mathdanisur last updated on 26/Jun/21
answer: (x/y)=(((√5)−1)/2) and (y/x)=(((√5)+1)/2) ?
$${answer}:\:\frac{{x}}{{y}}=\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}}\:{and}\:\frac{{y}}{{x}}=\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{2}}\:? \\ $$
Commented by liberty last updated on 26/Jun/21
from 4^k +10^k = 25^k divided by 25^k give ((2/5))^(2k) +((2/5))^k =1 where t =((2/5))^k ⇒t^2 +t−1 = 0 ⇒t =((−1 +(√5))/2) then (x/y)= (((√5)−1)/2) ← correct
$$\mathrm{from}\:\mathrm{4}^{\mathrm{k}} \:+\mathrm{10}^{\mathrm{k}} \:=\:\mathrm{25}^{\mathrm{k}} \:\mathrm{divided}\:\mathrm{by}\:\mathrm{25}^{\mathrm{k}} \\ $$$$\mathrm{give}\:\left(\frac{\mathrm{2}}{\mathrm{5}}\right)^{\mathrm{2k}} +\left(\frac{\mathrm{2}}{\mathrm{5}}\right)^{\mathrm{k}} =\mathrm{1}\:\mathrm{where}\:\mathrm{t}\:=\left(\frac{\mathrm{2}}{\mathrm{5}}\right)^{\mathrm{k}} \\ $$$$\Rightarrow\mathrm{t}^{\mathrm{2}} +\mathrm{t}−\mathrm{1}\:=\:\mathrm{0}\:\Rightarrow\mathrm{t}\:=\frac{−\mathrm{1}\:+\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$\mathrm{then}\:\frac{\mathrm{x}}{\mathrm{y}}=\:\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}}\:\leftarrow\:\mathrm{correct}\: \\ $$
Commented by haladu last updated on 26/Jun/21
Nice!
$$\boldsymbol{\mathrm{N}}{i}\boldsymbol{\mathrm{ce}}! \\ $$
Commented by mathdanisur last updated on 27/Jun/21
thanks Sir cool
$${thanks}\:{Sir}\:{cool} \\ $$
Answered by haladu last updated on 26/Jun/21
⇒ x = 4^h = ( 2^h )^2 ⇒ y = 10^h ⇔ y^2 = (2^h )^2 (5^h )^2 ⇒ x + y = ( 5^h )^2 y^2 = x ( x + y ) mutiply by (1/y^2 ) ( (x/y))^2 + ( (x/y)) −1 = 0 ⇒ (x/y) = ((−1 ±(√( 5)))/2)
$$\:\: \\ $$$$\:\:\:\Rightarrow\:\:\boldsymbol{\mathrm{x}}\:\:=\:\:\mathrm{4}^{\boldsymbol{\mathrm{h}}} \:=\:\left(\:\mathrm{2}^{\boldsymbol{\mathrm{h}}} \right)^{\mathrm{2}} \:\: \\ $$$$\:\Rightarrow\:\:\:\:\boldsymbol{\mathrm{y}}\:=\:\mathrm{10}^{\boldsymbol{\mathrm{h}}} \:\:\:\Leftrightarrow\:\boldsymbol{\mathrm{y}}^{\mathrm{2}} \:=\:\:\left(\mathrm{2}^{\boldsymbol{\mathrm{h}}} \right)^{\mathrm{2}} \left(\mathrm{5}^{\boldsymbol{\mathrm{h}}} \right)^{\mathrm{2}} \\ $$$$\:\: \\ $$$$\:\Rightarrow\:\boldsymbol{\mathrm{x}}\:\:+\:\boldsymbol{\mathrm{y}}\:\:=\:\:\left(\:\mathrm{5}\:^{\boldsymbol{\mathrm{h}}} \right)^{\mathrm{2}} \:\:\: \\ $$$$\:\:\: \\ $$$$\:\:\: \\ $$$$\:\:\:\:\:\boldsymbol{\mathrm{y}}^{\mathrm{2}} \:\:\:=\:\:\:\boldsymbol{\mathrm{x}}\:\:\left(\:\:\boldsymbol{\mathrm{x}}\:\:+\:\:\boldsymbol{\mathrm{y}}\:\right)\:\: \\ $$$$\:\:\:\: \\ $$$$\:\:\:\:\boldsymbol{\mathrm{mutiply}}\:\:\boldsymbol{\mathrm{by}}\:\:\:\frac{\mathrm{1}}{\boldsymbol{\mathrm{y}}^{\mathrm{2}} }\:\:\:\: \\ $$$$\:\:\: \\ $$$$\:\:\:\:\left(\:\frac{\boldsymbol{\mathrm{x}}}{\boldsymbol{\mathrm{y}}}\right)^{\mathrm{2}} +\:\left(\:\frac{\boldsymbol{\mathrm{x}}}{\boldsymbol{\mathrm{y}}}\right)\:−\mathrm{1}\:=\:\:\mathrm{0} \\ $$$$\:\: \\ $$$$\:\:\:\:\Rightarrow\:\:\:\:\frac{\boldsymbol{\mathrm{x}}}{\boldsymbol{\mathrm{y}}}\:\:=\:\:\:\frac{−\mathrm{1}\:\pm\sqrt{\:\mathrm{5}}}{\mathrm{2}} \\ $$$$\:\:\: \\ $$$$\:\: \\ $$
Commented by mathdanisur last updated on 27/Jun/21
thank you Sir cool
$${thank}\:{you}\:{Sir}\:{cool} \\ $$

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