Question Number 222026 by fantastic last updated on 15/Jun/25
$${If}\:\left(\mathrm{1}.\mathrm{234}\right)^{{a}} =\left(\mathrm{0}.\mathrm{1234}\right)^{{b}} =\mathrm{10}^{{c}} \\ $$$${prove}\:{that}\:\frac{\mathrm{1}}{{a}}−\frac{\mathrm{1}}{{c}}=\frac{\mathrm{1}}{{b}} \\ $$
Answered by som(math1967) last updated on 15/Jun/25
$$\left(\mathrm{1}.\mathrm{234}\right)^{{a}} =\left(\mathrm{0}.\mathrm{1234}\right)^{{b}} =\mathrm{10}^{{c}} ={k} \\ $$$$\:\mathrm{1}.\mathrm{234}={k}^{\frac{\mathrm{1}}{{a}}} \:\:\:\mathrm{0}.\mathrm{1234}={k}^{\frac{\mathrm{1}}{{b}}} \\ $$$$\:\mathrm{10}={k}^{\frac{\mathrm{1}}{{c}}} \\ $$$$\:\frac{\mathrm{1}.\mathrm{234}}{\mathrm{10}}=\frac{{k}^{\frac{\mathrm{1}}{{a}}} }{{k}^{\frac{\mathrm{1}}{{c}}} } \\ $$$$\mathrm{0}.\mathrm{1234}={k}^{\frac{\mathrm{1}}{{a}}−\frac{\mathrm{1}}{{c}}} \\ $$$$\:{k}^{\frac{\mathrm{1}}{{b}}} ={k}^{\frac{\mathrm{1}}{{a}}−\frac{\mathrm{1}}{{c}}} \\ $$$$\therefore\:\frac{\mathrm{1}}{{a}}−\frac{\mathrm{1}}{{c}}=\frac{\mathrm{1}}{{b}} \\ $$
Answered by Rasheed.Sindhi last updated on 15/Jun/25
$${If}\:\left(\mathrm{1}.\mathrm{234}\right)^{{a}} =\left(\mathrm{0}.\mathrm{1234}\right)^{{b}} =\mathrm{10}^{{c}} \\ $$$${prove}\:{that}\:\frac{\mathrm{1}}{{a}}−\frac{\mathrm{1}}{{c}}=\frac{\mathrm{1}}{{b}} \\ $$$$\left(\mathrm{1}.\mathrm{234}\right)^{{a}} =\left(\mathrm{0}.\mathrm{1234}\right)^{{b}} =\mathrm{10}^{{c}} \\ $$$$\left(\mathrm{0}.\mathrm{1234}×\mathrm{10}\right)^{{a}} =\left(\mathrm{0}.\mathrm{1234}\right)^{{b}} =\mathrm{10}^{{c}} \\ $$$$\left(\mathrm{0}.\mathrm{1234}\right)^{{a}} ×\left(\mathrm{10}\right)^{{a}} =\left(\mathrm{0}.\mathrm{1234}\right)^{{b}} =\mathrm{10}^{{c}} \\ $$$$\begin{cases}{\mathrm{0}.\mathrm{1234}=\mathrm{10}^{{c}/{b}} }\\{\mathrm{0}.\mathrm{1234}=\mathrm{10}^{\left({c}−{a}\right)/{a}} }\end{cases}\Rightarrow\frac{{c}}{{b}}=\frac{{c}−{a}}{{a}} \\ $$$$\frac{{c}}{{b}}=\frac{{c}−{a}}{{a}}\Rightarrow{ca}={bc}−{ab} \\ $$$$\frac{{ca}}{{abc}}=\frac{{bc}}{{abc}}−\frac{{ab}}{{abc}} \\ $$$$\frac{\mathrm{1}}{{b}}=\frac{\mathrm{1}}{{a}}−\frac{\mathrm{1}}{{c}} \\ $$$$\frac{\mathrm{1}}{{a}}−\frac{\mathrm{1}}{{c}}=\frac{\mathrm{1}}{{b}} \\ $$$${proved} \\ $$
Commented by fantastic last updated on 15/Jun/25
$$\mathbb{N}\mathrm{i}\mathbb{C}\epsilon! \\ $$
Commented by Rasheed.Sindhi last updated on 15/Jun/25
$$\underset{\leftarrow} {!\boldsymbol{\mathrm{s}}\Bbbk\boldsymbol{\mathrm{nah}}\mathbb{T}} \\ $$