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log-x-x-2-7-1-




Question Number 207559 by hardmath last updated on 18/May/24
log_x  (x^2  + 7)  ≤  1
$$\mathrm{log}_{\boldsymbol{\mathrm{x}}} \:\left(\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{7}\right)\:\:\leqslant\:\:\mathrm{1} \\ $$
Answered by mr W last updated on 19/May/24
log_x  (x^2 +7)−log_x  x≤0  log_x  (((x^2 +7)/x))≤0  ((log (((x^2 +7)/x)))/(log x))≤0  0<x<1 ∧ ((x^2 +7)/x)≥1  0<x<1 ∧ (x−(1/2))^2 +((27)/4) ≥0  ⇒0<x<1 ✓  or  1<x ∧ 0<((x^2 +7)/x)≤1  1<x ∧ (x−(1/2))^2 +((27)/4)≤0  ⇒no solution!
$$\mathrm{log}_{{x}} \:\left({x}^{\mathrm{2}} +\mathrm{7}\right)−\mathrm{log}_{{x}} \:{x}\leqslant\mathrm{0} \\ $$$$\mathrm{log}_{{x}} \:\left(\frac{{x}^{\mathrm{2}} +\mathrm{7}}{{x}}\right)\leqslant\mathrm{0} \\ $$$$\frac{\mathrm{log}\:\left(\frac{{x}^{\mathrm{2}} +\mathrm{7}}{{x}}\right)}{\mathrm{log}\:{x}}\leqslant\mathrm{0} \\ $$$$\mathrm{0}<{x}<\mathrm{1}\:\wedge\:\frac{{x}^{\mathrm{2}} +\mathrm{7}}{{x}}\geqslant\mathrm{1} \\ $$$$\mathrm{0}<{x}<\mathrm{1}\:\wedge\:\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{27}}{\mathrm{4}}\:\geqslant\mathrm{0} \\ $$$$\Rightarrow\mathrm{0}<{x}<\mathrm{1}\:\checkmark \\ $$$${or} \\ $$$$\mathrm{1}<{x}\:\wedge\:\mathrm{0}<\frac{{x}^{\mathrm{2}} +\mathrm{7}}{{x}}\leqslant\mathrm{1} \\ $$$$\mathrm{1}<{x}\:\wedge\:\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{27}}{\mathrm{4}}\leqslant\mathrm{0} \\ $$$$\Rightarrow{no}\:{solution}! \\ $$
Commented by hardmath last updated on 19/May/24
thank you professor
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{professor} \\ $$

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