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Question Number 56335 by Tawa1 last updated on 14/Mar/19

$$\mathrm{If}{\mathrm{b}}^{\mathrm{x}}={\left(\frac{\mathrm{b}}{\mathrm{k}}\right)}^{\mathrm{y}}={\mathrm{k}}^{\mathrm{m}}\mathrm{and}\mathrm{b}\ne 1$$$$\mathrm{show}\mathrm{that}:\frac{1}{\mathrm{x}}=\frac{1}{\mathrm{y}}=\frac{1}{\mathrm{m}}$$

Answered by tanmay.chaudhury50@gmail.com last updated on 14/Mar/19

$$b^x={\left(\frac{b}{k}\right)}^y=k^m=r\left(assumed\right)$$$$b^x=r\to b={\left(r\right)}^{\frac{1}{x}}$$$$k^m=r\to k={\left(r\right)}^{\frac{1}{m}}$$$${\left(\frac{b}{k}\right)}^y=r$$$${\left(\frac{r^{\frac{1}{x}}}{r^{\frac{1}{m}}}\right)}^y=r$$$$r^{\frac{1}{x}-\frac{1}{m}}=r^{\frac{1}{y}}$$$$\frac{1}{x}-\frac{1}{m}=\frac{1}{y}$$$$\frac{1}{x}+\frac{1}{y}=\frac{1}{m}$$

Commented by Tawa1 last updated on 14/Mar/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

Answered by $@ty@m last updated on 30/Mar/19

$$From{b}^x=k^{m}weget$$$$b=k^{\frac{m}{x}}--\left(1\right)$$$$From{\left(\frac{b}{k}\right)}^y=k^{m}weget$$$$b=k^{\frac{m+y}{y}}--\left(2\right)$$$$From\left(1\right)\mathrm{\&}\left(2\right),$$$$\frac{m}{x}=\frac{m+y}{y}$$$$\frac{m}{x}=\frac{m}{y}+1$$$$\frac{m}{x}=\frac{m}{y}+\frac{m}{m}$$$$\frac{m}{x}=m(\frac{1}{y}+\frac{1}{m})$$$$\frac{1}{x}=\frac{1}{y}+\frac{1}{m}$$$$$$

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