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

⌊ {2 \log}_{8} (x) + \frac{1}{3} ⌋ = \log_{4} (x) + \frac{1}{2}

Answered by MJS_new last updated on 03/Feb/23

lhs \in Z \Rightarrow x = 4^{n + \frac{1}{2}}; n \in Z

⌊ 2 \frac{(n + \frac{1}{2}) \log 4}{\log 8} + \frac{1}{3} ⌋ = n + 1

⌊ \frac{4}{3} n + 1 ⌋ = n + 1

n + 1 + ⌊ \frac{n}{3} ⌋ = n + 1

⌊ \frac{n}{3} ⌋ = 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

t h a n k s a l o t s i r . v e r y n i c e s o l u t i o n

Commented by MJS_new last updated on 03/Feb/23

{you}^{'} re welcome

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