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

$$\mathbf{solve}\mathbf{for}\mathit{x}\mathbf{and}\mathit{y}$$$$\left(\mathbf{i}\right)2^{\mathit{x}-1}.3^{\mathit{y}+1}=25$$$$\left(\mathbf{ii}\right)2^{\mathit{n}-1}.3^{\mathit{m}+1}=113$$

Commented by Amstrongmazoka last updated on 04/May/18

$$yousaidweshouldsolveforxandy$$$$yetweseenandminthequestion$$$$howaboutthat?$$

Answered by MJS last updated on 04/May/18

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

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

$$\mathbf{thanx}\mathbf{a}\mathbf{lot}\mathbf{sir}$$

Commented by MJS last updated on 04/May/18

$$...\mathrm{sorry}{\mathrm{there}}^{\prime }\mathrm{s}\mathrm{no}‘‘{\mathrm{beautiful}}^{″}\mathrm{solution}...$$

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

$$\mathrm{oky}$$

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

$$\mathrm{If}\mathrm{this}\mathrm{is}\mathrm{as}$$$$\left(\mathbf{i}\right)2^{\mathit{x}-1}.3^{\mathit{y}+1}=25$$$$\left(\mathbf{ii}\right)2^{\mathrm{y}-1}.3^{\mathrm{x}+1}=113$$$$\left(\mathrm{i}\right):2^{\mathrm{x}}.2^{-1}.3^{\mathrm{y}}.3=25$$$$\left(\mathrm{ii}\right):2^{\mathrm{y}}.2^{-1}.3^{\mathrm{x}}.3=113$$$$\left(\mathrm{i}\right)/\left(\mathrm{ii}\right):\frac{2^{\mathrm{x}}.2^{-1}.3^{\mathrm{y}}.3}{2^{\mathrm{y}}.2^{-1}.3^{\mathrm{x}}.3}=\frac{25}{113}$$$${\left(\frac{2}{3}\right)}^{\mathrm{x}}{\left(\frac{3}{2}\right)}^{\mathrm{y}}=\frac{25}{113}$$$${\left(\frac{2}{3}\right)}^{\mathrm{x}}{\left(\frac{2}{3}\right)}^{-\mathrm{y}}=\frac{25}{113}$$$${\left(\frac{2}{3}\right)}^{\mathrm{x}-\mathrm{y}}=\frac{25}{113}$$$$(\mathrm{x}-\mathrm{y})[\log 2-\log 3]=\log 25-\log 113$$$$\mathrm{x}-\mathrm{y}=\frac{\log 25-\log 113}{\log 2-\log 3}=\mathrm{a}\left(\mathrm{Say}\right)......\mathrm{I}$$$$\left(\mathrm{i}\right)\times \left(\mathrm{ii}\right):2^{\mathrm{x}+\mathrm{y}}.2^{-2}.3^{\mathrm{x}+\mathrm{y}}.3^2=25\times 113$$$${(2.3)}^{\mathrm{x}+\mathrm{y}}.\frac{9}{4}=25\times 113$$$$6^{\mathrm{x}+\mathrm{y}}=\frac{25\times 113\times 4}{9}$$$$\mathrm{x}+\mathrm{y}=\frac{\log 25+\log 113+\log 4-\log 9}{\log 6}=\mathrm{b}\left(\mathrm{say}\right)...\mathrm{II}$$$$\mathrm{I}+\mathrm{II}:\mathrm{x}=\frac{\mathrm{a}+\mathrm{b}}{2}$$$$\mathrm{II}-\mathrm{I}:\mathrm{y}=\frac{\mathrm{b}-\mathrm{a}}{2}$$

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

$$\mathbf{i}\mathbf{appreciate}\mathbf{this}!!!$$

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