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Question Number 18034 by chux last updated on 14/Jul/17

solve for a 5log _4 a+48log _a 4=(a/8)

$$\mathrm{solve}\:\mathrm{for}\:\mathrm{a} \\ $$$$ \\ $$$$\mathrm{5log}\:_{\mathrm{4}} \mathrm{a}+\mathrm{48log}\:_{\mathrm{a}} \mathrm{4}=\frac{\mathrm{a}}{\mathrm{8}} \\ $$

Answered by 433 last updated on 14/Jul/17

5log _4 a+48((log _4 4)/(log _4 a))=(a/8) 5log _4 a+((48)/(log _4 a))=(a/8) log _4 a=x ⇔a=4^x 5x+((48)/x)=(4^x /8) 5x+((48)/x)=2^(2x−3) (1) x=4 ⇒ 20+((48)/4)=2^5 If x_0 <0 was solution 5x_0 +((48)/x_0 )<0 & 2^(2x_0 −3) >0 x>0 f(x)=5x+((48)/x)−2^(2x−3) f′(x)=5−((48)/x^2 )−2^(2x−3) ×ln 2×2 0<x<3 ((48)/x^2 )+2^(2x−1) ×ln 2>((48)/x^2 )+((ln 2)/2)>((48)/9)+((ln 2)/2)>5 x>3 ((48)/x^2 )+2^(2x−1) ×ln 2>((48)/x^2 )+32×ln 2>5 f′(x)<0 (1) has ≤1 solution x=4 ⇒ a=4^4

$$ \\ $$$$ \\ $$$$\mathrm{5log}\:_{\mathrm{4}} {a}+\mathrm{48}\frac{\mathrm{log}\:_{\mathrm{4}} \mathrm{4}}{\mathrm{log}\:_{\mathrm{4}} {a}}=\frac{{a}}{\mathrm{8}} \\ $$$$\mathrm{5log}\:_{\mathrm{4}} {a}+\frac{\mathrm{48}}{\mathrm{log}\:_{\mathrm{4}} {a}}=\frac{{a}}{\mathrm{8}} \\ $$$$\mathrm{log}\:_{\mathrm{4}} {a}={x}\:\Leftrightarrow{a}=\mathrm{4}^{{x}} \\ $$$$\mathrm{5}{x}+\frac{\mathrm{48}}{{x}}=\frac{\mathrm{4}^{{x}} }{\mathrm{8}} \\ $$$$\mathrm{5}{x}+\frac{\mathrm{48}}{{x}}=\mathrm{2}^{\mathrm{2}{x}−\mathrm{3}} \:\:\left(\mathrm{1}\right) \\ $$$${x}=\mathrm{4}\:\Rightarrow\:\mathrm{20}+\frac{\mathrm{48}}{\mathrm{4}}=\mathrm{2}^{\mathrm{5}} \: \\ $$$${If}\:{x}_{\mathrm{0}} <\mathrm{0}\:{was}\:{solution} \\ $$$$\mathrm{5}{x}_{\mathrm{0}} +\frac{\mathrm{48}}{{x}_{\mathrm{0}} }<\mathrm{0}\:\&\:\mathrm{2}^{\mathrm{2}{x}_{\mathrm{0}} −\mathrm{3}} >\mathrm{0} \\ $$$${x}>\mathrm{0} \\ $$$${f}\left({x}\right)=\mathrm{5}{x}+\frac{\mathrm{48}}{{x}}−\mathrm{2}^{\mathrm{2}{x}−\mathrm{3}} \\ $$$${f}'\left({x}\right)=\mathrm{5}−\frac{\mathrm{48}}{{x}^{\mathrm{2}} }−\mathrm{2}^{\mathrm{2}{x}−\mathrm{3}} ×\mathrm{ln}\:\mathrm{2}×\mathrm{2} \\ $$$$\mathrm{0}<{x}<\mathrm{3} \\ $$$$\frac{\mathrm{48}}{{x}^{\mathrm{2}} }+\mathrm{2}^{\mathrm{2}{x}−\mathrm{1}} ×\mathrm{ln}\:\mathrm{2}>\frac{\mathrm{48}}{{x}^{\mathrm{2}} }+\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{2}}>\frac{\mathrm{48}}{\mathrm{9}}+\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{2}}>\mathrm{5} \\ $$$${x}>\mathrm{3} \\ $$$$\frac{\mathrm{48}}{{x}^{\mathrm{2}} }+\mathrm{2}^{\mathrm{2}{x}−\mathrm{1}} ×\mathrm{ln}\:\mathrm{2}>\frac{\mathrm{48}}{{x}^{\mathrm{2}} }+\mathrm{32}×\mathrm{ln}\:\mathrm{2}>\mathrm{5} \\ $$$${f}'\left({x}\right)<\mathrm{0} \\ $$$$\left(\mathrm{1}\right)\:{has}\:\leqslant\mathrm{1}\:{solution} \\ $$$${x}=\mathrm{4}\:\Rightarrow\:{a}=\mathrm{4}^{\mathrm{4}} \\ $$

Answered by ajfour last updated on 14/Jul/17

a=256 let a=4^x , then log _4 a=x ⇒ 5x+((48)/x)= (4^(x−1) /2) ⇒ x=4 (trial and error with derivative test) ⇒ a=4^4 =256 (only one solution).

$$\:\mathrm{a}=\mathrm{256} \\ $$$$\mathrm{let}\:\mathrm{a}=\mathrm{4}^{\boldsymbol{\mathrm{x}}} \:,\:\mathrm{then}\:\mathrm{log}\:_{\mathrm{4}} \mathrm{a}=\mathrm{x} \\ $$$$\Rightarrow\:\:\mathrm{5x}+\frac{\mathrm{48}}{\mathrm{x}}=\:\frac{\mathrm{4}^{\boldsymbol{\mathrm{x}}−\mathrm{1}} }{\mathrm{2}}\: \\ $$$$\Rightarrow\:\:\mathrm{x}=\mathrm{4}\:\:\:\:\:\:\:\:\left(\mathrm{trial}\:\mathrm{and}\:\mathrm{error}\:\mathrm{with}\right. \\ $$$$\left.\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{derivative}\:\mathrm{test}\right) \\ $$$$\Rightarrow\:\mathrm{a}=\mathrm{4}^{\mathrm{4}} =\mathrm{256}\:\:\:\:\left(\mathrm{only}\:\mathrm{one}\:\mathrm{solution}\right). \\ $$$$ \\ $$

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