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Question Number 34328 by mondodotto@gmail.com last updated on 04/May/18

solve for x and y (i) 2^(x−1) .3^(y+1) =25 (ii)2^(n−1) .3^(m+1) =113

$$\boldsymbol{\mathrm{solve}}\:\boldsymbol{\mathrm{for}}\:\boldsymbol{{x}}\:\boldsymbol{\mathrm{and}}\:\boldsymbol{{y}} \\ $$$$\left(\boldsymbol{\mathrm{i}}\right)\:\mathrm{2}^{\boldsymbol{{x}}−\mathrm{1}} .\mathrm{3}^{\boldsymbol{{y}}+\mathrm{1}} =\mathrm{25} \\ $$$$\left(\boldsymbol{\mathrm{ii}}\right)\mathrm{2}^{\boldsymbol{{n}}−\mathrm{1}} .\mathrm{3}^{\boldsymbol{{m}}+\mathrm{1}} =\mathrm{113} \\ $$

Commented by Amstrongmazoka last updated on 04/May/18

you said we should solve for x and y yet we see n and m in the question how about that?

$${you}\:{said}\:{we}\:{should}\:{solve}\:{for}\:{x}\:{and}\:{y} \\ $$$${yet}\:{we}\:{see}\:{n}\:{and}\:{m}\:{in}\:{the}\:{question} \\ $$$${how}\:{about}\:{that}? \\ $$

Answered by MJS last updated on 04/May/18

(i) 3^(y+1) =25×2^(1−x) (y+1)ln 3=ln 25+(1−x)ln 2 y=−((ln 2)/(ln 3))x+((ln 2−ln 3+2ln 5)/(ln 3)) (ii) 2^(y−1) 3^(x+1) =113 2^(y−1) =113×3^(−x−1) (y−1)ln 2=ln 113−(x+1)ln 3 y=−((ln 3)/(ln 2))x+((ln 2−ln 3+ln 113)/(ln 2)) −((ln 2)/(ln 3))x+((ln 2−ln 3+2ln 5)/(ln 3))=−((ln 3)/(ln 2))x+((ln 2−ln 3+ln 113)/(ln 2)) x=((ln^2 2−2ln 2×(ln 3−ln 5)+ln 3×(ln 3−ln 113))/(ln^2 2−ln^2 3)) y=((ln^2 2−ln 2×(2ln 3−ln 113)+ln 3×(ln 3−2ln 5))/(ln^2 2−ln^2 3))

$$\left(\mathrm{i}\right)\:\mathrm{3}^{{y}+\mathrm{1}} =\mathrm{25}×\mathrm{2}^{\mathrm{1}−{x}} \\ $$$$\left({y}+\mathrm{1}\right)\mathrm{ln}\:\mathrm{3}=\mathrm{ln}\:\mathrm{25}+\left(\mathrm{1}−{x}\right)\mathrm{ln}\:\mathrm{2} \\ $$$${y}=−\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{ln}\:\mathrm{3}}{x}+\frac{\mathrm{ln}\:\mathrm{2}−\mathrm{ln}\:\mathrm{3}+\mathrm{2ln}\:\mathrm{5}}{\mathrm{ln}\:\mathrm{3}} \\ $$$$ \\ $$$$\left(\mathrm{ii}\right)\:\mathrm{2}^{{y}−\mathrm{1}} \mathrm{3}^{{x}+\mathrm{1}} =\mathrm{113} \\ $$$$\mathrm{2}^{{y}−\mathrm{1}} =\mathrm{113}×\mathrm{3}^{−{x}−\mathrm{1}} \\ $$$$\left({y}−\mathrm{1}\right)\mathrm{ln}\:\mathrm{2}=\mathrm{ln}\:\mathrm{113}−\left({x}+\mathrm{1}\right)\mathrm{ln}\:\mathrm{3} \\ $$$${y}=−\frac{\mathrm{ln}\:\mathrm{3}}{\mathrm{ln}\:\mathrm{2}}{x}+\frac{\mathrm{ln}\:\mathrm{2}−\mathrm{ln}\:\mathrm{3}+\mathrm{ln}\:\mathrm{113}}{\mathrm{ln}\:\mathrm{2}} \\ $$$$ \\ $$$$−\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{ln}\:\mathrm{3}}{x}+\frac{\mathrm{ln}\:\mathrm{2}−\mathrm{ln}\:\mathrm{3}+\mathrm{2ln}\:\mathrm{5}}{\mathrm{ln}\:\mathrm{3}}=−\frac{\mathrm{ln}\:\mathrm{3}}{\mathrm{ln}\:\mathrm{2}}{x}+\frac{\mathrm{ln}\:\mathrm{2}−\mathrm{ln}\:\mathrm{3}+\mathrm{ln}\:\mathrm{113}}{\mathrm{ln}\:\mathrm{2}} \\ $$$${x}=\frac{\mathrm{ln}^{\mathrm{2}} \:\mathrm{2}−\mathrm{2ln}\:\mathrm{2}×\left(\mathrm{ln}\:\mathrm{3}−\mathrm{ln}\:\mathrm{5}\right)+\mathrm{ln}\:\mathrm{3}×\left(\mathrm{ln}\:\mathrm{3}−\mathrm{ln}\:\mathrm{113}\right)}{\mathrm{ln}^{\mathrm{2}} \:\mathrm{2}−\mathrm{ln}^{\mathrm{2}} \:\mathrm{3}} \\ $$$${y}=\frac{\mathrm{ln}^{\mathrm{2}} \:\mathrm{2}−\mathrm{ln}\:\mathrm{2}×\left(\mathrm{2ln}\:\mathrm{3}−\mathrm{ln}\:\mathrm{113}\right)+\mathrm{ln}\:\mathrm{3}×\left(\mathrm{ln}\:\mathrm{3}−\mathrm{2ln}\:\mathrm{5}\right)}{\mathrm{ln}^{\mathrm{2}} \:\mathrm{2}−\mathrm{ln}^{\mathrm{2}} \:\mathrm{3}} \\ $$$$ \\ $$

Commented by mondodotto@gmail.com last updated on 04/May/18

thanx a lot sir

$$\boldsymbol{\mathrm{thanx}}\:\boldsymbol{\mathrm{a}}\:\boldsymbol{\mathrm{lot}}\:\boldsymbol{\mathrm{sir}} \\ $$

Commented by MJS last updated on 04/May/18

...sorry there′s no “beautiful” solution...

$$...\mathrm{sorry}\:\mathrm{there}'\mathrm{s}\:\mathrm{no}\:``\mathrm{beautiful}''\:\mathrm{solution}... \\ $$

Commented by mondodotto@gmail.com last updated on 05/May/18

oky

$$\mathrm{oky} \\ $$

Answered by Rasheed.Sindhi last updated on 06/May/18

If this is as (i) 2^(x−1) .3^(y+1) =25 (ii)2^(y−1) .3^(x+1) =113 (i): 2^x .2^(−1) .3^y .3=25 (ii): 2^y .2^(−1) .3^x .3=113 (i)/(ii): (( 2^x .2^(−1) .3^y .3)/(2^y .2^(−1) .3^x .3))=((25)/(113)) ((2/3))^x ((3/2))^y =((25)/(113)) ((2/3))^x ((2/3))^(−y) =((25)/(113)) ((2/3))^(x−y) =((25)/(113)) (x−y)[log2−log3]=log25−log113 x−y=((log25−log113)/(log2−log3))=a (Say)......I (i)×(ii):2^(x+y) .2^(−2) .3^(x+y) .3^2 =25×113 (2.3)^(x+y) .(9/4)=25×113 6^(x+y) =((25×113×4)/9) x+y=((log25+log113+log4−log9)/(log6))=b (say)...II I+II: x=((a+b)/2) II−I: y=((b−a)/2)

$$\mathrm{If}\:\mathrm{this}\:\mathrm{is}\:\mathrm{as} \\ $$$$\left(\boldsymbol{\mathrm{i}}\right)\:\mathrm{2}^{\boldsymbol{{x}}−\mathrm{1}} .\mathrm{3}^{\boldsymbol{{y}}+\mathrm{1}} =\mathrm{25} \\ $$$$\left(\boldsymbol{\mathrm{ii}}\right)\mathrm{2}^{\mathrm{y}−\mathrm{1}} .\mathrm{3}^{\mathrm{x}+\mathrm{1}} =\mathrm{113} \\ $$$$\:\:\left(\mathrm{i}\right):\:\:\mathrm{2}^{\mathrm{x}} .\mathrm{2}^{−\mathrm{1}} .\mathrm{3}^{\mathrm{y}} .\mathrm{3}=\mathrm{25} \\ $$$$\:\:\left(\mathrm{ii}\right):\:\mathrm{2}^{\mathrm{y}} .\mathrm{2}^{−\mathrm{1}} .\mathrm{3}^{\mathrm{x}} .\mathrm{3}=\mathrm{113} \\ $$$$\:\left(\mathrm{i}\right)/\left(\mathrm{ii}\right):\:\frac{\:\mathrm{2}^{\mathrm{x}} .\mathrm{2}^{−\mathrm{1}} .\mathrm{3}^{\mathrm{y}} .\mathrm{3}}{\mathrm{2}^{\mathrm{y}} .\mathrm{2}^{−\mathrm{1}} .\mathrm{3}^{\mathrm{x}} .\mathrm{3}}=\frac{\mathrm{25}}{\mathrm{113}} \\ $$$$\:\:\:\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{x}} \left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{y}} =\frac{\mathrm{25}}{\mathrm{113}} \\ $$$$\:\:\:\:\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{x}} \left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{−\mathrm{y}} =\frac{\mathrm{25}}{\mathrm{113}} \\ $$$$\:\:\:\:\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{x}−\mathrm{y}} =\frac{\mathrm{25}}{\mathrm{113}} \\ $$$$\:\:\:\left(\mathrm{x}−\mathrm{y}\right)\left[\mathrm{log2}−\mathrm{log3}\right]=\mathrm{log25}−\mathrm{log113} \\ $$$$\:\:\:\:\:\:\mathrm{x}−\mathrm{y}=\frac{\mathrm{log25}−\mathrm{log113}}{\mathrm{log2}−\mathrm{log3}}=\mathrm{a}\:\left(\mathrm{Say}\right)......\mathrm{I} \\ $$$$\left(\mathrm{i}\right)×\left(\mathrm{ii}\right):\mathrm{2}^{\mathrm{x}+\mathrm{y}} .\mathrm{2}^{−\mathrm{2}} .\mathrm{3}^{\mathrm{x}+\mathrm{y}} .\mathrm{3}^{\mathrm{2}} =\mathrm{25}×\mathrm{113} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{2}.\mathrm{3}\right)^{\mathrm{x}+\mathrm{y}} .\frac{\mathrm{9}}{\mathrm{4}}=\mathrm{25}×\mathrm{113} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{6}^{\mathrm{x}+\mathrm{y}} =\frac{\mathrm{25}×\mathrm{113}×\mathrm{4}}{\mathrm{9}} \\ $$$$\:\:\:\:\:\:\mathrm{x}+\mathrm{y}=\frac{\mathrm{log25}+\mathrm{log113}+\mathrm{log4}−\mathrm{log9}}{\mathrm{log6}}=\mathrm{b}\:\left(\mathrm{say}\right)...\mathrm{II} \\ $$$$\mathrm{I}+\mathrm{II}:\:\:\mathrm{x}=\frac{\mathrm{a}+\mathrm{b}}{\mathrm{2}} \\ $$$$\mathrm{II}−\mathrm{I}:\:\:\mathrm{y}=\frac{\mathrm{b}−\mathrm{a}}{\mathrm{2}} \\ $$

Commented by mondodotto@gmail.com last updated on 04/May/18

i appreciate this!!!

$$\boldsymbol{\mathrm{i}}\:\boldsymbol{\mathrm{appreciate}}\:\boldsymbol{\mathrm{this}}!!! \\ $$

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