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solve-in-R-2log-8-x-1-3-log-4-x-1-2-




Question Number 186300 by mnjuly1970 last updated on 03/Feb/23
          solve  in   R       ⌊  2log_( 8) (x) + (1/3) ⌋ = log_( 4) (x )+ (1/2)
$$\:\:\: \\ $$$$\:\:\:\:\:\mathrm{solve}\:\:\mathrm{in}\:\:\:\mathbb{R} \\ $$$$ \\ $$$$\:\:\:\lfloor\:\:\mathrm{2log}_{\:\mathrm{8}} \left({x}\right)\:+\:\frac{\mathrm{1}}{\mathrm{3}}\:\rfloor\:=\:\mathrm{log}_{\:\mathrm{4}} \left({x}\:\right)+\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$ \\ $$
Answered by MJS_new last updated on 03/Feb/23
lhs ∈Z ⇒ x=4^(n+(1/2)) ; n∈Z  ⌊2(((n+(1/2))log 4)/(log 8))+(1/3)⌋=n+1  ⌊(4/3)n+1⌋=n+1  n+1+⌊(n/3)⌋=n+1  ⌊(n/3)⌋=0  n=0, 1, 2  x=4^(1/2) , 4^(3/2) , 4^(5/2) =2, 8, 32
$$\mathrm{lhs}\:\in\mathbb{Z}\:\Rightarrow\:{x}=\mathrm{4}^{{n}+\frac{\mathrm{1}}{\mathrm{2}}} ;\:{n}\in\mathbb{Z} \\ $$$$\lfloor\mathrm{2}\frac{\left({n}+\frac{\mathrm{1}}{\mathrm{2}}\right)\mathrm{log}\:\mathrm{4}}{\mathrm{log}\:\mathrm{8}}+\frac{\mathrm{1}}{\mathrm{3}}\rfloor={n}+\mathrm{1} \\ $$$$\lfloor\frac{\mathrm{4}}{\mathrm{3}}{n}+\mathrm{1}\rfloor={n}+\mathrm{1} \\ $$$${n}+\mathrm{1}+\lfloor\frac{{n}}{\mathrm{3}}\rfloor={n}+\mathrm{1} \\ $$$$\lfloor\frac{{n}}{\mathrm{3}}\rfloor=\mathrm{0} \\ $$$${n}=\mathrm{0},\:\mathrm{1},\:\mathrm{2} \\ $$$${x}=\mathrm{4}^{\mathrm{1}/\mathrm{2}} ,\:\mathrm{4}^{\mathrm{3}/\mathrm{2}} ,\:\mathrm{4}^{\mathrm{5}/\mathrm{2}} =\mathrm{2},\:\mathrm{8},\:\mathrm{32} \\ $$
Commented by mnjuly1970 last updated on 03/Feb/23
thanks alot  sir . very nice solution
$${thanks}\:{alot}\:\:{sir}\:.\:{very}\:{nice}\:{solution} \\ $$
Commented by MJS_new last updated on 03/Feb/23
you′re welcome
$$\mathrm{you}'\mathrm{re}\:\mathrm{welcome} \\ $$

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