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Question-12284

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Question Number 12284 by tawa last updated on 17/Apr/17
Answered by mrW1 last updated on 18/Apr/17
let t=(1/(2x)) log_5 (5^(1/x) +125)=log_5 6+1+(1/(2x)) ⇒ log_5 (5^(2t) +5^3 )=log_5 6+(1+t) log_5 (5^(2t) +5^3 )−log_5 6−log_5 5^(1+t) =0 log_5 ((5^(2t) +5^3 )/(5^(1+t) ×6))=0 ((5^(2t) +5^3 )/(5^(1+t) ×6))=1 ((5^(t−1) +5^(2−t) )/6)=1 5^(t−1) +5^(2−t) =6=5^0 +5^1 possible solutions: t−1=0 and 2−t=1 ⇒t=1⇒x=(1/2) t−1=1 and 2−t=0 ⇒t=2 ⇒x=(1/4)
$${let}\:{t}=\frac{\mathrm{1}}{\mathrm{2}{x}} \\ $$$$\mathrm{log}_{\mathrm{5}} \:\left(\mathrm{5}^{\frac{\mathrm{1}}{{x}}} +\mathrm{125}\right)=\mathrm{log}_{\mathrm{5}} \:\mathrm{6}+\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}{x}}\:\Rightarrow \\ $$$$\mathrm{log}_{\mathrm{5}} \:\left(\mathrm{5}^{\mathrm{2}{t}} +\mathrm{5}^{\mathrm{3}} \right)=\mathrm{log}_{\mathrm{5}} \:\mathrm{6}+\left(\mathrm{1}+{t}\right) \\ $$$$\mathrm{log}_{\mathrm{5}} \:\left(\mathrm{5}^{\mathrm{2}{t}} +\mathrm{5}^{\mathrm{3}} \right)−\mathrm{log}_{\mathrm{5}} \:\mathrm{6}−\mathrm{log}_{\mathrm{5}} \:\mathrm{5}^{\mathrm{1}+{t}} =\mathrm{0} \\ $$$$\mathrm{log}_{\mathrm{5}} \:\frac{\mathrm{5}^{\mathrm{2}{t}} +\mathrm{5}^{\mathrm{3}} }{\mathrm{5}^{\mathrm{1}+{t}} ×\mathrm{6}}=\mathrm{0} \\ $$$$\frac{\mathrm{5}^{\mathrm{2}{t}} +\mathrm{5}^{\mathrm{3}} }{\mathrm{5}^{\mathrm{1}+{t}} ×\mathrm{6}}=\mathrm{1} \\ $$$$\frac{\mathrm{5}^{{t}−\mathrm{1}} +\mathrm{5}^{\mathrm{2}−{t}} }{\mathrm{6}}=\mathrm{1} \\ $$$$\mathrm{5}^{{t}−\mathrm{1}} +\mathrm{5}^{\mathrm{2}−{t}} =\mathrm{6}=\mathrm{5}^{\mathrm{0}} +\mathrm{5}^{\mathrm{1}} \\ $$$$ \\ $$$${possible}\:{solutions}: \\ $$$${t}−\mathrm{1}=\mathrm{0}\:{and}\:\mathrm{2}−{t}=\mathrm{1}\:\Rightarrow{t}=\mathrm{1}\Rightarrow{x}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${t}−\mathrm{1}=\mathrm{1}\:{and}\:\mathrm{2}−{t}=\mathrm{0}\:\Rightarrow{t}=\mathrm{2}\:\Rightarrow{x}=\frac{\mathrm{1}}{\mathrm{4}} \\ $$
Commented by tawa last updated on 18/Apr/17
i really appreciate. God bless you sir.
$$\mathrm{i}\:\mathrm{really}\:\mathrm{appreciate}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$
Answered by frank ntulah last updated on 18/Apr/17
⇒log_5 (5^(1/x) +125)=log_5 6+log_5 5+(1/(2x)) ⇒log_5 (((5^(1/x) +125)/(30)))=(1/(2x)) ⇒(((5^(1/x) +125)/(30)))=5^(1/(2x)) ⇒let y=5^(((1/x))) ⇒(((y+125)/(30)))=y^(1/2) ⇒ y+125=30y^(((1/2))) ⇒y_1 ^(((1/2))) =25 y_2 ^(((1/2))) =5 ⇒1=4x ⇒1=2x x_1 =0.25 x_2 =0.5
$$\Rightarrow\mathrm{log}_{\mathrm{5}} \left(\mathrm{5}^{\frac{\mathrm{1}}{\mathrm{x}}} +\mathrm{125}\right)=\mathrm{log}_{\mathrm{5}} \mathrm{6}+\mathrm{log}_{\mathrm{5}} \mathrm{5}+\frac{\mathrm{1}}{\mathrm{2x}} \\ $$$$\Rightarrow\mathrm{log}_{\mathrm{5}} \left(\frac{\mathrm{5}^{\frac{\mathrm{1}}{\mathrm{x}}} +\mathrm{125}}{\mathrm{30}}\right)=\frac{\mathrm{1}}{\mathrm{2x}} \\ $$$$\Rightarrow\left(\frac{\mathrm{5}^{\frac{\mathrm{1}}{\mathrm{x}}} +\mathrm{125}}{\mathrm{30}}\right)=\mathrm{5}^{\frac{\mathrm{1}}{\mathrm{2x}}} \\ $$$$\Rightarrow\mathrm{let}\:\mathrm{y}=\mathrm{5}^{\left(\frac{\mathrm{1}}{\mathrm{x}}\right)} \\ $$$$\Rightarrow\left(\frac{\mathrm{y}+\mathrm{125}}{\mathrm{30}}\right)=\mathrm{y}^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$\Rightarrow\:\mathrm{y}+\mathrm{125}=\mathrm{30y}^{\left(\frac{\mathrm{1}}{\mathrm{2}}\right)} \\ $$$$\Rightarrow\mathrm{y}_{\mathrm{1}} ^{\left(\frac{\mathrm{1}}{\mathrm{2}}\right)} =\mathrm{25} \\ $$$$\mathrm{y}_{\mathrm{2}} ^{\left(\frac{\mathrm{1}}{\mathrm{2}}\right)} =\mathrm{5} \\ $$$$\Rightarrow\mathrm{1}=\mathrm{4x} \\ $$$$\Rightarrow\mathrm{1}=\mathrm{2x} \\ $$$$\mathrm{x}_{\mathrm{1}} =\mathrm{0}.\mathrm{25} \\ $$$$\mathrm{x}_{\mathrm{2}} =\mathrm{0}.\mathrm{5} \\ $$
Commented by tawa last updated on 18/Apr/17
I really appreciate. God bless you sir.
$$\mathrm{I}\:\mathrm{really}\:\mathrm{appreciate}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

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