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Question Number 186300 by mnjuly1970 last updated on 03/Feb/23

$$$$$$\mathrm{solve}\mathrm{in}\mathbb{R}$$$$$$$$\lfloor {2\log }_8\left(x\right)+\frac{1}{3}\rfloor ={\log }_4\left(x\right)+\frac{1}{2}$$$$$$

Answered by MJS_new last updated on 03/Feb/23

$$\mathrm{lhs}\in \mathbb{Z}\Rightarrow x=4^{n+\frac{1}{2}};n\in \mathbb{Z}$$$$\lfloor 2\frac{(n+\frac{1}{2})\log 4}{\log 8}+\frac{1}{3}\rfloor =n+1$$$$\lfloor \frac{4}{3}n+1\rfloor =n+1$$$$n+1+\lfloor \frac{n}{3}\rfloor =n+1$$$$\lfloor \frac{n}{3}\rfloor =0$$$$n=0,1,2$$$$x=4^{1/2},4^{3/2},4^{5/2}=2,8,32$$

Commented by mnjuly1970 last updated on 03/Feb/23

$$thanksalotsir.verynicesolution$$

Commented by MJS_new last updated on 03/Feb/23

$${\mathrm{you}}^{\prime }\mathrm{re}\mathrm{welcome}$$

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