Question Number 224538 by MirHasibulHossain last updated on 17/Sep/25
$$\mathrm{32}^{\mathrm{4r}^{\mathrm{2}} −\mathrm{8}} =\mathrm{1}\:\:\:\:\mathrm{then}\:\mathrm{find}\:\mathrm{r}=? \\ $$
Answered by fantastic last updated on 17/Sep/25
$$\left(\mathrm{4}{r}^{\mathrm{2}} −\mathrm{8}\right)=\mathrm{log}\:_{\mathrm{32}} \mathrm{1}=\mathrm{0} \\ $$$$\mathrm{4}{r}^{\mathrm{2}} =\mathrm{8} \\ $$$${r}=\pm\sqrt{\mathrm{2}} \\ $$
Answered by Rasheed.Sindhi last updated on 17/Sep/25
$$\mathrm{32}^{\mathrm{4}{r}^{\mathrm{2}} −\mathrm{8}} =\mathrm{1} \\ $$$$\mathrm{32}^{\mathrm{4}{r}^{\mathrm{2}} −\mathrm{8}} =\mathrm{32}^{\mathrm{0}} \\ $$$$\mathrm{4}{r}^{\mathrm{2}} −\mathrm{8}=\mathrm{0} \\ $$$${r}^{\mathrm{2}} =\frac{\mathrm{8}}{\mathrm{4}}=\mathrm{2} \\ $$$${r}=\pm\sqrt{\mathrm{2}}\: \\ $$