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Question Number 222125 by hardmath last updated on 18/Jun/25
If: log_3 (5^x + (1/5^x ) + 7) ⇒ min = ?
$$\mathrm{If}:\:\:\:\mathrm{log}_{\mathrm{3}} \left(\mathrm{5}^{\boldsymbol{\mathrm{x}}} \:+\:\frac{\mathrm{1}}{\mathrm{5}^{\boldsymbol{\mathrm{x}}} }\:+\:\mathrm{7}\right)\:\:\:\Rightarrow\:\:\:\mathrm{min}\:=\:? \\ $$
Commented by mr W last updated on 18/Jun/25
you mean if y=log_3 (5^x +(1/5^x )+7), find y_(min) =?
$${you}\:{mean} \\ $$$${if}\:{y}=\mathrm{log}_{\mathrm{3}} \:\left(\mathrm{5}^{{x}} +\frac{\mathrm{1}}{\mathrm{5}^{{x}} }+\mathrm{7}\right),\:{find}\:{y}_{{min}} =? \\ $$
Commented by hardmath last updated on 18/Jun/25
yes dear professor
$$\mathrm{yes}\:\mathrm{dear}\:\mathrm{professor} \\ $$
Answered by gregori last updated on 19/Jun/25
let h(x)= 5^x +(1/5^x ) +7 ⇒min h(x)= 9 when 5^x =1 or x=0 y_(min) = log _3 (5^x +(1/5^x ) +7 )= log _3 (9)=2
$$\:{let}\:{h}\left({x}\right)=\:\mathrm{5}^{{x}} +\frac{\mathrm{1}}{\mathrm{5}^{{x}} }\:+\mathrm{7}\:\Rightarrow{min}\:{h}\left({x}\right)=\:\mathrm{9} \\ $$$$\:\:{when}\:\mathrm{5}^{{x}} =\mathrm{1}\:{or}\:{x}=\mathrm{0} \\ $$$$\:{y}_{{min}} \:=\:\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{5}^{{x}} +\frac{\mathrm{1}}{\mathrm{5}^{{x}} }\:+\mathrm{7}\:\right)=\:\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{9}\right)=\mathrm{2} \\ $$$$ \\ $$
Commented by hardmath last updated on 19/Jun/25
thank you, but when 5^x =1 or x=0 how?
$$\mathrm{thank}\:\mathrm{you},\:\mathrm{but} \\ $$$$\mathrm{when}\:\mathrm{5}^{\boldsymbol{\mathrm{x}}} =\mathrm{1}\:\mathrm{or}\:\mathrm{x}=\mathrm{0}\:\mathrm{how}? \\ $$
Answered by mr W last updated on 19/Jun/25
y=log_3 (5^x +(1/5^x )+7) ≥log_3 (2×(√(5^x ×(1/5^x )))+7)=log_3 9=2 ⇒y_(min) =2 (at x=0)
$${y}=\mathrm{log}_{\mathrm{3}} \:\left(\mathrm{5}^{{x}} +\frac{\mathrm{1}}{\mathrm{5}^{{x}} }+\mathrm{7}\right) \\ $$$$\:\:\:\geqslant\mathrm{log}_{\mathrm{3}} \:\left(\mathrm{2}×\sqrt{\mathrm{5}^{{x}} ×\frac{\mathrm{1}}{\mathrm{5}^{{x}} }}+\mathrm{7}\right)=\mathrm{log}_{\mathrm{3}} \:\mathrm{9}=\mathrm{2} \\ $$$$\Rightarrow{y}_{{min}} =\mathrm{2}\:\:\:\:\left({at}\:{x}=\mathrm{0}\right) \\ $$
Commented by hardmath last updated on 18/Jun/25
thank you dear professor, but ... (2 ∙ (√(5^x ∙ (1/5^x ))) + 7) how?
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{dear}\:\mathrm{professor},\:\mathrm{but} \\ $$$$…\:\left(\mathrm{2}\:\centerdot\:\sqrt{\mathrm{5}^{\boldsymbol{\mathrm{x}}} \:\centerdot\:\frac{\mathrm{1}}{\mathrm{5}^{\boldsymbol{\mathrm{x}}} }}\:+\:\mathrm{7}\right)\:\:\mathrm{how}? \\ $$
Commented by mr W last updated on 19/Jun/25
for positive values a, b we have ((a+b)/2)≥(√(ab)) this is the so−called AM≥GM. (arithmetic mean ≥ geometeric mean) in other form: a+b≥2(√(ab)) the equalty holds when a=b. therefore: 5^x , (1/5^x ) >0 ⇒5^x +(1/5^x )≥2(√(5^x ×(1/5^x )))=2 the equalty holds when 5^x =(1/5^x ), i.e. (5^x )^2 =1, i.e. x=0 the proof for AM≥GM: a+b−2(√(ab)) =((√a))^2 −2((√a))((√b))+((√b))^2 =((√a)−(√b))^2 ≥0 ⇒a+b≥2(√(ab)) or ((a+b)/2)≥(√(ab)) the general form for AM≥GM is: if a_1 , a_2 , ..., a_n >0, then a_1 +a_2 +...+a_n ≥n((a_1 a_2 ...a_n ))^(1/n)
$${for}\:{positive}\:{values}\:{a},\:{b}\:{we}\:{have} \\ $$$$\frac{{a}+{b}}{\mathrm{2}}\geqslant\sqrt{{ab}} \\ $$$${this}\:{is}\:{the}\:{so}−{called}\:{AM}\geqslant{GM}. \\ $$$$\left({arithmetic}\:{mean}\:\geqslant\:{geometeric}\:{mean}\right) \\ $$$${in}\:{other}\:{form}:\:{a}+{b}\geqslant\mathrm{2}\sqrt{{ab}} \\ $$$${the}\:{equalty}\:{holds}\:{when}\:{a}={b}. \\ $$$${therefore}: \\ $$$$\mathrm{5}^{{x}} ,\:\frac{\mathrm{1}}{\mathrm{5}^{{x}} }\:>\mathrm{0} \\ $$$$\Rightarrow\mathrm{5}^{{x}} +\frac{\mathrm{1}}{\mathrm{5}^{{x}} }\geqslant\mathrm{2}\sqrt{\mathrm{5}^{{x}} ×\frac{\mathrm{1}}{\mathrm{5}^{{x}} }}=\mathrm{2} \\ $$$${the}\:{equalty}\:{holds}\:{when} \\ $$$$\mathrm{5}^{{x}} =\frac{\mathrm{1}}{\mathrm{5}^{{x}} },\:{i}.{e}.\:\left(\mathrm{5}^{{x}} \right)^{\mathrm{2}} =\mathrm{1},\:{i}.{e}.\:{x}=\mathrm{0} \\ $$$$ \\ $$$${the}\:{proof}\:{for}\:{AM}\geqslant{GM}: \\ $$$${a}+{b}−\mathrm{2}\sqrt{{ab}} \\ $$$$=\left(\sqrt{{a}}\right)^{\mathrm{2}} −\mathrm{2}\left(\sqrt{{a}}\right)\left(\sqrt{{b}}\right)+\left(\sqrt{{b}}\right)^{\mathrm{2}} \\ $$$$=\left(\sqrt{{a}}−\sqrt{{b}}\right)^{\mathrm{2}} \geqslant\mathrm{0} \\ $$$$\Rightarrow{a}+{b}\geqslant\mathrm{2}\sqrt{{ab}}\:{or}\:\frac{{a}+{b}}{\mathrm{2}}\geqslant\sqrt{{ab}} \\ $$$$ \\ $$$${the}\:{general}\:{form}\:{for}\:{AM}\geqslant{GM}\:{is}: \\ $$$${if}\:{a}_{\mathrm{1}} ,\:{a}_{\mathrm{2}} ,\:…,\:{a}_{{n}} >\mathrm{0},\:{then} \\ $$$${a}_{\mathrm{1}} +{a}_{\mathrm{2}} +…+{a}_{{n}} \geqslant{n}\sqrt[{{n}}]{{a}_{\mathrm{1}} {a}_{\mathrm{2}} …{a}_{{n}} } \\ $$
Commented by hardmath last updated on 19/Jun/25
thank you dear professors
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{dear}\:\mathrm{professors} \\ $$
Answered by fantastic last updated on 19/Jun/25
log _3 (5^x +(1/5^x )+7) is an increasing function to find the minimum we have to find (5^x +(1/5^x )+7)_(min) So (d/dx)(5^x +(1/5^x )+7)=0 (d/dx)(5^x +(1/5^x )+7)=5^x ln (5)+(−5^(−x) ln (5))+0 So 5^x ln (5)−5^(−x) ln (5)=o or ln (5)(5^x −5^(−x) )=0 So 5^x −5^(−x) =0[∵ln (5)≠0] or 5^x =5^(−x) So x=0 So when x=0 the value of (5^x +(1/5^x )+7)is minimum When (5^x +(1/5^x )+7)is min log _3 (5^x +(1/5^x )+7)is also min entering x=o 5^0 +(1/5^0 )+7=1+1+7=9 So the ninimum value is log _3 9=2 FINAL ANSWER THE MINIMUM VALUE OF log _3 (5^x +(1/5^x )+7) IS 2
$$\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{5}^{{x}} +\frac{\mathrm{1}}{\mathrm{5}^{{x}} }+\mathrm{7}\right)\:{is}\:{an}\:{increasing}\:{function} \\ $$$${to}\:{find}\:\:{the}\:{minimum}\:{we}\:{have}\:{to}\:{find}\:\left(\mathrm{5}^{{x}} +\frac{\mathrm{1}}{\mathrm{5}^{{x}} }+\mathrm{7}\right)_{{min}} \\ $$$${So}\:\frac{{d}}{{dx}}\left(\mathrm{5}^{{x}} +\frac{\mathrm{1}}{\mathrm{5}^{{x}} }+\mathrm{7}\right)=\mathrm{0} \\ $$$$\frac{{d}}{{dx}}\left(\mathrm{5}^{{x}} +\frac{\mathrm{1}}{\mathrm{5}^{{x}} }+\mathrm{7}\right)=\mathrm{5}^{{x}} \mathrm{ln}\:\left(\mathrm{5}\right)+\left(−\mathrm{5}^{−{x}} \mathrm{ln}\:\left(\mathrm{5}\right)\right)+\mathrm{0} \\ $$$${So}\:\mathrm{5}^{{x}} \mathrm{ln}\:\left(\mathrm{5}\right)−\mathrm{5}^{−{x}} \mathrm{ln}\:\left(\mathrm{5}\right)={o} \\ $$$${or}\:\mathrm{ln}\:\left(\mathrm{5}\right)\left(\mathrm{5}^{{x}} −\mathrm{5}^{−{x}} \right)=\mathrm{0} \\ $$$${So}\:\mathrm{5}^{{x}} −\mathrm{5}^{−{x}} =\mathrm{0}\left[\because\mathrm{ln}\:\left(\mathrm{5}\right)\neq\mathrm{0}\right] \\ $$$${or}\:\mathrm{5}^{{x}} =\mathrm{5}^{−{x}} \\ $$$${So}\:{x}=\mathrm{0} \\ $$$${So}\:{when}\:{x}=\mathrm{0}\:{the}\:{value}\:{of}\:\left(\mathrm{5}^{{x}} +\frac{\mathrm{1}}{\mathrm{5}^{{x}} }+\mathrm{7}\right){is}\:{minimum} \\ $$$${When}\:\left(\mathrm{5}^{{x}} +\frac{\mathrm{1}}{\mathrm{5}^{{x}} }+\mathrm{7}\right){is}\:{min}\:\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{5}^{{x}} +\frac{\mathrm{1}}{\mathrm{5}^{{x}} }+\mathrm{7}\right){is}\:{also}\:{min} \\ $$$${entering}\:{x}={o} \\ $$$$\mathrm{5}^{\mathrm{0}} +\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{0}} }+\mathrm{7}=\mathrm{1}+\mathrm{1}+\mathrm{7}=\mathrm{9} \\ $$$${So}\:{the}\:{ninimum}\:{value}\:{is}\:\mathrm{log}\:_{\mathrm{3}} \mathrm{9}=\mathrm{2} \\ $$$${FINAL}\:{ANSWER}\:\:{THE}\:{MINIMUM}\:{VALUE}\:{OF}\:\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{5}^{{x}} +\frac{\mathrm{1}}{\mathrm{5}^{{x}} }+\mathrm{7}\right)\:{IS}\:\mathrm{2} \\ $$
Answered by fantastic last updated on 19/Jun/25
We can also use AM−GM rule which states a,b>0 then ((a+b)/2)≥(√(ab)) So in the case of 5^x +(1/5^x ) it is (((5^x +(1/5^x ))/2))≥(√(5^x ×(1/5^x ))) or (((5^x +(1/5^x ))/2))≥(√1) or (((5^x +(1/5^x ))/2))≥1 or 5^x +(1/5^x )≥2 This shows that (5^x +(1/5^x ))_(min) =2 So (log _3 (5^x +(1/5^x )+7))_(min) =log _3 (2+7)=log _3 9=2✓
$${We}\:{can}\:{also}\:{use}\:{AM}−{GM}\:{rule}\:{which}\:{states} \\ $$$${a},{b}>\mathrm{0}\:{then}\:\frac{{a}+{b}}{\mathrm{2}}\geqslant\sqrt{{ab}} \\ $$$${So}\:{in}\:{the}\:{case}\:{of}\:\mathrm{5}^{{x}} +\frac{\mathrm{1}}{\mathrm{5}^{{x}} }\:{it}\:{is} \\ $$$$\left(\frac{\mathrm{5}^{{x}} +\frac{\mathrm{1}}{\mathrm{5}^{{x}} }}{\mathrm{2}}\right)\geqslant\sqrt{\mathrm{5}^{{x}} ×\frac{\mathrm{1}}{\mathrm{5}^{{x}} }} \\ $$$${or}\:\left(\frac{\mathrm{5}^{{x}} +\frac{\mathrm{1}}{\mathrm{5}^{{x}} }}{\mathrm{2}}\right)\geqslant\sqrt{\mathrm{1}} \\ $$$${or}\:\left(\frac{\mathrm{5}^{{x}} +\frac{\mathrm{1}}{\mathrm{5}^{{x}} }}{\mathrm{2}}\right)\geqslant\mathrm{1} \\ $$$${or}\:\mathrm{5}^{{x}} +\frac{\mathrm{1}}{\mathrm{5}^{{x}} }\geqslant\mathrm{2} \\ $$$${This}\:{shows}\:{that}\:\left(\mathrm{5}^{{x}} +\frac{\mathrm{1}}{\mathrm{5}^{{x}} }\right)_{{min}} =\mathrm{2} \\ $$$${So}\:\left(\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{5}^{{x}} +\frac{\mathrm{1}}{\mathrm{5}^{{x}} }+\mathrm{7}\right)\right)_{{min}} =\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{2}+\mathrm{7}\right)=\mathrm{log}\:_{\mathrm{3}} \mathrm{9}=\mathrm{2}\checkmark \\ $$
Commented by hardmath last updated on 19/Jun/25
thank you dear professors
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{dear}\:\mathrm{professors} \\ $$

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