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Question Number 144563 by mathdanisur last updated on 26/Jun/21

$${\mathit{l}\mathit{o}\mathit{g}}_4\left(\mathit{x}\right)=\mathit{l}\mathit{o}\mathit{g}\left(\mathit{y}\right)={\mathit{l}\mathit{o}\mathit{g}}_{25}(\mathit{x}+\mathit{y})$$$$Find\frac{x}{y}=?$$

Answered by imjagoll last updated on 26/Jun/21

$$\mathrm{let}\log {}_4\left(\mathrm{x}\right)=\log {}_{10}\left(\mathrm{y}\right)=\log {}_{25}(\mathrm{x}+\mathrm{y})=\mathrm{k}$$$$\mathrm{then}\{\begin{array}{l}\mathrm{x}=4^{\mathrm{k}}\\ \mathrm{y}={10}^{\mathrm{k}}\\ \mathrm{x}+\mathrm{y}={25}^{\mathrm{k}}\end{array}$$$$\iff 4^{\mathrm{k}}+{10}^{\mathrm{k}}={25}^{\mathrm{k}}$$$$\Rightarrow 1+{\left(\frac{5}{2}\right)}^{\mathrm{k}}={\left(\frac{5}{2}\right)}^{2\mathrm{k}}$$$$\Rightarrow {\mathrm{t}}^2-\mathrm{t}-1=0;\mathrm{where}\mathrm{t}={\left(\frac{5}{2}\right)}^{\mathrm{k}}$$$$\Rightarrow \mathrm{t}=\frac{1+\sqrt{5}}{2}$$$$\Rightarrow \mathrm{we}\mathrm{find}\frac{\mathrm{x}}{\mathrm{y}}=\frac{4^{\mathrm{k}}}{{10}^{\mathrm{k}}}={\left(\frac{2}{5}\right)}^{\mathrm{k}}=\frac{2}{\sqrt{5}+1}$$$$\frac{\mathrm{x}}{\mathrm{y}}=\frac{2(\sqrt{5}-1)}{4}=\frac{\sqrt{5}-1}{2}$$

Commented by mathdanisur last updated on 26/Jun/21

$$alotcoolthanksSir,answer\frac{1}{2}(\sqrt{5}+1)$$

Commented by imjagoll last updated on 26/Jun/21

$$\mathrm{wrong}.\mathrm{if}\frac{\mathrm{y}}{\mathrm{x}}=\frac{\sqrt{5}+1}{2}\mathrm{true}$$

Commented by mathdanisur last updated on 26/Jun/21

$$answer:\frac{x}{y}=\frac{\sqrt{5}-1}{2}and\frac{y}{x}=\frac{\sqrt{5}+1}{2}?$$

Commented by liberty last updated on 26/Jun/21

$$\mathrm{from}{4}^{\mathrm{k}}+{10}^{\mathrm{k}}={25}^{\mathrm{k}}\mathrm{divided}\mathrm{by}{25}^{\mathrm{k}}$$$$\mathrm{give}{\left(\frac{2}{5}\right)}^{2\mathrm{k}}+{\left(\frac{2}{5}\right)}^{\mathrm{k}}=1\mathrm{where}\mathrm{t}={\left(\frac{2}{5}\right)}^{\mathrm{k}}$$$$\Rightarrow {\mathrm{t}}^2+\mathrm{t}-1=0\Rightarrow \mathrm{t}=\frac{-1+\sqrt{5}}{2}$$$$\mathrm{then}\frac{\mathrm{x}}{\mathrm{y}}=\frac{\sqrt{5}-1}{2}\leftarrow \mathrm{correct}$$

Commented by haladu last updated on 26/Jun/21

$$\mathbf{N}i\mathbf{ce}!$$

Commented by mathdanisur last updated on 27/Jun/21

$$thanksSircool$$

Answered by haladu last updated on 26/Jun/21

$$$$$$\Rightarrow \mathbf{x}=4^{\mathbf{h}}={\left(2^{\mathbf{h}}\right)}^2$$$$\Rightarrow \mathbf{y}={10}^{\mathbf{h}}\iff {\mathbf{y}}^2={\left(2^{\mathbf{h}}\right)}^2{\left(5^{\mathbf{h}}\right)}^2$$$$$$$$\Rightarrow \mathbf{x}+\mathbf{y}={\left(5{}^{\mathbf{h}}\right)}^2$$$$$$$$$$$${\mathbf{y}}^2=\mathbf{x}(\mathbf{x}+\mathbf{y})$$$$$$$$\mathbf{mutiply}\mathbf{by}\frac{1}{{\mathbf{y}}^2}$$$$$$$${\left(\frac{\mathbf{x}}{\mathbf{y}}\right)}^2+\left(\frac{\mathbf{x}}{\mathbf{y}}\right)-1=0$$$$$$$$\Rightarrow \frac{\mathbf{x}}{\mathbf{y}}=\frac{-1\pm \sqrt{5}}{2}$$$$$$$$$$

Commented by mathdanisur last updated on 27/Jun/21

$$thankyouSircool$$

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