Логарифмы. Решите неравенства. №28, 29. Created by Sasha9290. matematika-ru

28. lgx≥1ОДЗ: x>01=lg10lgx≥lg10основание логарифма а=10, => знак неравенства не меняем.x∈(0;10]1+ 2+ 3+4+5+6+ 7+8+ 9+ 10=5529. log₀,₅x≥-2ОДЗ: x>0log₀,₅x ≥log₀,₅4основание логарифма а=0,5. 0<0,5<1, => знак неравенства меняемx∈(0;4]1*2*3*4=24

Логарифмы. Решите неравенства. №28, 29...

Sasha9290 @Sasha9290

July 2022 1 9 Report

Логарифмы. Решите неравенства. №28, 29

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kirichekov

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28. lgx≥1

ОДЗ: x>0
1=lg10

lgx≥lg10
основание логарифма а=10, => знак неравенства не меняем.
$\left \{ {{x\ \textgreater \ 0} \atop {x \leq 10}} \right.$
x∈(0;10]
1+ 2+ 3+4+5+6+ 7+8+ 9+ 10=55

29. log₀,₅x≥-2
ОДЗ: x>0
$-2= log_{0,5}0,5^{-2}= log_{ \frac{1}{2} } ( \frac{1}{2} )^{-2} = log_{0,5}4$
log₀,₅x ≥log₀,₅4
основание логарифма а=0,5. 0<0,5<1, => знак неравенства меняем
$\left \{ {{x\ \textgreater \ 0} \atop {x \leq 4}} \right.$
x∈(0;4]
1*2*3*4=24

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