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Find-x-48log-a-4-5log-4-a-a-8-




Question Number 41797 by Tawa1 last updated on 12/Aug/18
Find x:        48log_a 4  +  5log_4 a  =  (a/8)
$$\mathrm{Find}\:\mathrm{x}:\:\:\:\:\:\:\:\:\mathrm{48log}_{\mathrm{a}} \mathrm{4}\:\:+\:\:\mathrm{5log}_{\mathrm{4}} \mathrm{a}\:\:=\:\:\frac{\mathrm{a}}{\mathrm{8}} \\ $$
Answered by alex041103 last updated on 13/Aug/18
log_b a=((lna)/(lnb)) and log_a b=((lnb)/(lna))  ⇒log_4 a=(1/(log_a 4))  ⇒let log_4 a=t  ⇒48t^2 +5−(a/8)t=0  ⇒t_(1,2) =(((a/8)±(√((a^2 /(64))−960)))/(96))=log_4 a  ....W function...
$${log}_{{b}} {a}=\frac{{lna}}{{lnb}}\:{and}\:{log}_{{a}} {b}=\frac{{lnb}}{{lna}} \\ $$$$\Rightarrow{log}_{\mathrm{4}} {a}=\frac{\mathrm{1}}{{log}_{{a}} \mathrm{4}} \\ $$$$\Rightarrow{let}\:{log}_{\mathrm{4}} {a}={t} \\ $$$$\Rightarrow\mathrm{48}{t}^{\mathrm{2}} +\mathrm{5}−\frac{{a}}{\mathrm{8}}{t}=\mathrm{0} \\ $$$$\Rightarrow{t}_{\mathrm{1},\mathrm{2}} =\frac{\frac{{a}}{\mathrm{8}}\pm\sqrt{\frac{{a}^{\mathrm{2}} }{\mathrm{64}}−\mathrm{960}}}{\mathrm{96}}={log}_{\mathrm{4}} {a} \\ $$$$….{W}\:{function}… \\ $$
Commented by Tawa1 last updated on 13/Aug/18
God bless you sir
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$
Commented by MrW3 last updated on 13/Aug/18
working is not correct sir!  solution for  ax^2 +bx+c=0  is x=((−b±(√(b^2 −4ac)))/(2a)) only   when a,b,c are constants, i.e. when  they are indepentent from x!  but in 48t^2 +5−(a/8)t=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  (x^2 −1)^2 +2x(x^2 −1)+1=0  you can not solve it like this:  x^2 −1=((−2x±(√(4x^2 −4)))/2)=−x±(√(x^2 −1))
$${working}\:{is}\:{not}\:{correct}\:{sir}! \\ $$$${solution}\:{for} \\ $$$${ax}^{\mathrm{2}} +{bx}+{c}=\mathrm{0} \\ $$$${is}\:{x}=\frac{−{b}\pm\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}}{\mathrm{2}{a}}\:{only}\: \\ $$$${when}\:{a},{b},{c}\:{are}\:{constants},\:{i}.{e}.\:{when} \\ $$$${they}\:{are}\:{indepentent}\:{from}\:{x}! \\ $$$${but}\:{in}\:\mathrm{48}{t}^{\mathrm{2}} +\mathrm{5}−\frac{{a}}{\mathrm{8}}{t}=\mathrm{0}\:{t}\:{and}\:{a}\:{are}\:{dependent} \\ $$$${from}\:{each}\:{other},\:{so}\:{you}\:{may}\:{not}\:{apply} \\ $$$${the}\:{formula}\:{for}\:{quadratic}\:{equation}. \\ $$$${if}\:{I}\:{have}\:{an}\:{eqn}.\:{like}\:{this} \\ $$$$\left({x}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} +\mathrm{2}{x}\left({x}^{\mathrm{2}} −\mathrm{1}\right)+\mathrm{1}=\mathrm{0} \\ $$$${you}\:{can}\:{not}\:{solve}\:{it}\:{like}\:{this}: \\ $$$${x}^{\mathrm{2}} −\mathrm{1}=\frac{−\mathrm{2}{x}\pm\sqrt{\mathrm{4}{x}^{\mathrm{2}} −\mathrm{4}}}{\mathrm{2}}=−{x}\pm\sqrt{{x}^{\mathrm{2}} −\mathrm{1}} \\ $$
Commented by Tawa1 last updated on 15/Aug/18
God bless you sir.
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$
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...
$${That}\:{is}\:{not}\:{a}\:{problem}.{If}\:{you}\:{have}\:{seen} \\ $$$${the}\:{derivation}\:{of}\:{the}\:{quadratic}\:{equation}, \\ $$$${it}\:{is}\:{made}\:{by}\:{doing}\:{valid}\:{algebraic} \\ $$$${manipulation}.{In}\:{our}\:{case},\:{we}\:{are} \\ $$$${in}\:{fact}\:{just}\:{transforming}\:{the}\:{equation} \\ $$$${into}\:{some}\:{other}\:{equation},\:{we}\:{aren}'{t}\:{solving}\:{it}. \\ $$$${We}\:{just}\:{use}\:{the}\:{quadratic}\:{equation} \\ $$$${to}\:{transform}\:{it}\:{into}\:{some}\:{other}\:{equation} \\ $$$${which}\:{is}\:{more}\:{familiar}… \\ $$

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