Question Number 19194 by mondodotto@gmail.com last updated on 06/Aug/17
Answered by allizzwell23 last updated on 06/Aug/17
$$\:\:\:\:\:\left(\frac{\mathrm{25}}{\mathrm{4}}\:\right)^{\mathrm{3}\left(\mathrm{x}−\mathrm{1}\right)} =\:\left(\frac{\mathrm{4}}{\mathrm{10}}\right)^{\mathrm{3}\left(\mathrm{5}−\mathrm{x}\right)} \\ $$$$\:\:\:\:\:\left(\frac{\mathrm{5}}{\mathrm{2}}\:\right)^{\mathrm{2}\left(\mathrm{x}−\mathrm{1}\right)} =\:\left(\frac{\mathrm{2}}{\mathrm{5}}\right)^{\left(\mathrm{5}−\mathrm{x}\right)} \\ $$$$\:\:\:\:\:\left(\frac{\mathrm{5}}{\mathrm{2}}\:\right)^{\mathrm{2}\left(\mathrm{x}−\mathrm{1}\right)} =\:\left(\frac{\mathrm{5}}{\mathrm{2}}\right)^{−\left(\mathrm{5}−\mathrm{x}\right)} \\ $$$$\:\:\Rightarrow\:\:\:\mathrm{2}\left(\mathrm{x}−\mathrm{1}\right)\:=\:\mathrm{x}−\mathrm{5} \\ $$$$\:\:\therefore\:\:\:\:\:\:\:\mathrm{x}\:=\:−\mathrm{3} \\ $$$$ \\ $$$$ \\ $$
Answered by ajfour last updated on 06/Aug/17
$$\left(\mathrm{6}.\mathrm{25}\right)^{\left(\mathrm{3x}−\mathrm{3}\right)} =\left(\mathrm{0}.\mathrm{064}\right)^{\left(\mathrm{5}−\mathrm{x}\right)} \\ $$$$\Rightarrow\:\:\:\:\left(\frac{\mathrm{25}}{\mathrm{4}}\right)^{\mathrm{3x}−\mathrm{3}} =\left(\frac{\mathrm{8}}{\mathrm{25}}\right)^{\mathrm{5}−\mathrm{x}} \\ $$$$\Rightarrow\:\:\:\:\left(\frac{\mathrm{5}}{\mathrm{2}}\right)^{\mathrm{6x}−\mathrm{6}} =\left(\frac{\mathrm{2}}{\mathrm{5}}\right)^{\mathrm{15}−\mathrm{3x}} \\ $$$$\Rightarrow\:\:\:\:\:\left(\frac{\mathrm{5}}{\mathrm{2}}\right)^{\mathrm{6x}−\mathrm{6}} =\left(\frac{\mathrm{5}}{\mathrm{2}}\right)^{\mathrm{3x}−\mathrm{15}} \\ $$$$\mathrm{or}\:\:\:\:\:\:\:\mathrm{6x}−\mathrm{6}=\mathrm{3x}−\mathrm{15} \\ $$$$\:\:\:\:\:\:\:\:\:\Rightarrow\:\:\:\mathrm{x}=−\mathrm{3}\:. \\ $$