Question Number 41797 by Tawa1 last updated on 12/Aug/18
$$\mathrm{Find}\:\mathrm{x}:\:\:\:\:\:\:\:\:\mathrm{48log}_{\mathrm{a}} \mathrm{4}\:\:+\:\:\mathrm{5log}_{\mathrm{4}} \mathrm{a}\:\:=\:\:\frac{\mathrm{a}}{\mathrm{8}} \\ $$
Answered by alex041103 last updated on 13/Aug/18
$${log}_{{b}} {a}=\frac{{lna}}{{lnb}}\:{and}\:{log}_{{a}} {b}=\frac{{lnb}}{{lna}} \\ $$$$\Rightarrow{log}_{\mathrm{4}} {a}=\frac{\mathrm{1}}{{log}_{{a}} \mathrm{4}} \\ $$$$\Rightarrow{let}\:{log}_{\mathrm{4}} {a}={t} \\ $$$$\Rightarrow\mathrm{48}{t}^{\mathrm{2}} +\mathrm{5}−\frac{{a}}{\mathrm{8}}{t}=\mathrm{0} \\ $$$$\Rightarrow{t}_{\mathrm{1},\mathrm{2}} =\frac{\frac{{a}}{\mathrm{8}}\pm\sqrt{\frac{{a}^{\mathrm{2}} }{\mathrm{64}}−\mathrm{960}}}{\mathrm{96}}={log}_{\mathrm{4}} {a} \\ $$$$....{W}\:{function}... \\ $$
Commented by Tawa1 last updated on 13/Aug/18
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$
Commented by MrW3 last updated on 13/Aug/18
$${working}\:{is}\:{not}\:{correct}\:{sir}! \\ $$$${solution}\:{for} \\ $$$${ax}^{\mathrm{2}} +{bx}+{c}=\mathrm{0} \\ $$$${is}\:{x}=\frac{−{b}\pm\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}}{\mathrm{2}{a}}\:{only}\: \\ $$$${when}\:{a},{b},{c}\:{are}\:{constants},\:{i}.{e}.\:{when} \\ $$$${they}\:{are}\:{indepentent}\:{from}\:{x}! \\ $$$${but}\:{in}\:\mathrm{48}{t}^{\mathrm{2}} +\mathrm{5}−\frac{{a}}{\mathrm{8}}{t}=\mathrm{0}\:{t}\:{and}\:{a}\:{are}\:{dependent} \\ $$$${from}\:{each}\:{other},\:{so}\:{you}\:{may}\:{not}\:{apply} \\ $$$${the}\:{formula}\:{for}\:{quadratic}\:{equation}. \\ $$$${if}\:{I}\:{have}\:{an}\:{eqn}.\:{like}\:{this} \\ $$$$\left({x}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} +\mathrm{2}{x}\left({x}^{\mathrm{2}} −\mathrm{1}\right)+\mathrm{1}=\mathrm{0} \\ $$$${you}\:{can}\:{not}\:{solve}\:{it}\:{like}\:{this}: \\ $$$${x}^{\mathrm{2}} −\mathrm{1}=\frac{−\mathrm{2}{x}\pm\sqrt{\mathrm{4}{x}^{\mathrm{2}} −\mathrm{4}}}{\mathrm{2}}=−{x}\pm\sqrt{{x}^{\mathrm{2}} −\mathrm{1}} \\ $$
Commented by Tawa1 last updated on 15/Aug/18
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$
Commented by alex041103 last updated on 18/Aug/18
$${That}\:{is}\:{not}\:{a}\:{problem}.{If}\:{you}\:{have}\:{seen} \\ $$$${the}\:{derivation}\:{of}\:{the}\:{quadratic}\:{equation}, \\ $$$${it}\:{is}\:{made}\:{by}\:{doing}\:{valid}\:{algebraic} \\ $$$${manipulation}.{In}\:{our}\:{case},\:{we}\:{are} \\ $$$${in}\:{fact}\:{just}\:{transforming}\:{the}\:{equation} \\ $$$${into}\:{some}\:{other}\:{equation},\:{we}\:{aren}'{t}\:{solving}\:{it}. \\ $$$${We}\:{just}\:{use}\:{the}\:{quadratic}\:{equation} \\ $$$${to}\:{transform}\:{it}\:{into}\:{some}\:{other}\:{equation} \\ $$$${which}\:{is}\:{more}\:{familiar}... \\ $$