Question Number 56534 by Gulay last updated on 18/Mar/19
Commented by Gulay last updated on 18/Mar/19
$$\mathrm{sir}\:\mathrm{could}\:\mathrm{you}\:\mathrm{help}\:\mathrm{me} \\ $$
Answered by MJS last updated on 18/Mar/19
$$\mathrm{log}_{{b}} \:{p}\:=\mathrm{log}_{{b}} \:{q}\:\Leftrightarrow\:{p}={q} \\ $$$$\mathrm{2}{x}−\mathrm{3}={x}+\mathrm{4}\:\Rightarrow\:{x}=\mathrm{7} \\ $$
Commented by Gulay last updated on 18/Mar/19
$$\mathrm{thanks}\:\mathrm{sir} \\ $$