Question Number 1355 by Rasheed Ahmad last updated on 25/Jul/15
$$\mathrm{8}^{{log}\:\left(\mathrm{12}{x}+\mathrm{1}\right)} =\mathrm{4}^{{log}\:\mathrm{27}} \:\:\:,{solve}\:{for}\:{x}. \\ $$
Answered by Yugi last updated on 25/Jul/15
$${Rewriting}\:{the}\:{above}\:{equation}\:{in}\:{base}\:\mathrm{2}\:{gives} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}^{\mathrm{3}{log}\left(\mathrm{12}{x}+\mathrm{1}\right)} =\mathrm{2}^{\mathrm{2}{log}\mathrm{27}} ……….\left(\mathrm{1}\right) \\ $$$${Since}\:{the}\:{bases}\:{are}\:{the}\:{same},\:{we}\:{can}\:{equate} \\ $$$${the}\:{indices}\:{on}\:{both}\:{sides}\:{of}\:\:\left(\mathrm{1}\right). \\ $$$$\therefore\:\mathrm{3}{log}\left(\mathrm{12}{x}+\mathrm{1}\right)=\mathrm{2}{log}\mathrm{3}^{\mathrm{3}} \\ $$$$\mathrm{3}{log}\left(\mathrm{12}{x}+\mathrm{1}\right)=\mathrm{6}{log}\mathrm{3} \\ $$$$\boldsymbol{\div}\mathrm{3}:\:{log}\left(\mathrm{12}{x}+\mathrm{1}\right)={log}\mathrm{9} \\ $$$${Hence}\:\mathrm{10}^{{log}\left(\mathrm{12}{x}+\mathrm{1}\right)} =\mathrm{10}^{{log}\mathrm{9}} \\ $$$$\therefore\:\mathrm{12}{x}+\mathrm{1}=\mathrm{9}\:\Rightarrow{x}=\frac{\mathrm{2}}{\mathrm{3}}\: \\ $$
Commented by Rasheed Soomro last updated on 25/Jul/15
$${Thanks}.\:{Appreciation}\:{for}\:{your}\:{approach}. \\ $$