Question Number 8934 by j.masanja06@gmail.com last updated on 06/Nov/16
$$\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\:\mathrm{x}. \\ $$$$\mathrm{3}^{\mathrm{x}+\mathrm{1}} =\mathrm{2}^{\mathrm{x}+\mathrm{2}} \\ $$
Answered by Rasheed Soomro last updated on 08/Nov/16
$$\mathrm{3}^{\mathrm{x}+\mathrm{1}} =\mathrm{2}^{\mathrm{x}+\mathrm{2}} \\ $$$$\mathrm{3}^{\mathrm{x}} .\mathrm{3}=\mathrm{2}^{\mathrm{x}} .\mathrm{2}^{\mathrm{2}} \\ $$$$\frac{\mathrm{3}^{\mathrm{x}} }{\mathrm{2}^{\mathrm{x}} }=\frac{\mathrm{4}}{\mathrm{3}} \\ $$$$\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{x}} =\frac{\mathrm{4}}{\mathrm{3}} \\ $$$$\mathrm{xlog}\left(\frac{\mathrm{3}}{\mathrm{2}}\right)=\mathrm{log}\left(\frac{\mathrm{4}}{\mathrm{3}}\right) \\ $$$$\mathrm{x}=\frac{\mathrm{log4}−\mathrm{log3}}{\mathrm{log3}−\mathrm{log2}}\approx\mathrm{0}.\mathrm{7095} \\ $$$$\mathrm{x}\approx\mathrm{0}.\mathrm{7095} \\ $$
Commented by j.masanja06@gmail.com last updated on 07/Nov/16
$$\mathrm{sir}\:\mathrm{remember}\:\mathrm{that}\:\mathrm{2}^{\mathrm{x}+\mathrm{2}} \:\mathrm{and}\:\mathrm{not}\:\mathrm{2}^{\mathrm{x}+\mathrm{1}} \\ $$
Commented by Rasheed Soomro last updated on 07/Nov/16
$$\mathrm{Sorry}\:\mathrm{for}\:\mathrm{misunderstanding}! \\ $$
Commented by j.masanja06@gmail.com last updated on 07/Nov/16
$$\mathrm{ok}\:\mathrm{sir}! \\ $$
Commented by j.masanja06@gmail.com last updated on 07/Nov/16
$$\mathrm{nicely}\:\mathrm{sir} \\ $$