What is log_{3}(log_{10}(3x))<log_{3}(81) ?

The solution to log_{3}(log_{10}(3x))<log_{3}(81) is 1/3 <x<(10^{81})/3

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log_{3}(log_{10}(3x))<log_{3}(81)

lo g_{3} (lo g_{10} (3 x)) < lo g_{3} (81)

\frac{1}{3} < x < \frac{1 0 ^{81}}{3}

Interval Notation

(\frac{1}{3}, \frac{1 0 ^{81}}{3})

Decimal Notation

0.33333 \dots < x < 3.33333 E 80

Solution steps

lo g_{3} (lo g_{10} (3 x)) < lo g_{3} (81)

Apply log rules

lo g_{10} (3 x) < 81 and lo g_{10} (3 x) > 0

lo g_{10} (3 x) < 81 : 0 < x < \frac{1 0 ^{81}}{3}

lo g_{10} (3 x) > 0 : x > \frac{1}{3}

Combine the intervals

0 < x < \frac{1 0 ^{81}}{3} and x > \frac{1}{3}

Merge Overlapping Intervals

\frac{1}{3} < x < \frac{1 0 ^{81}}{3}

lo g_{2} (x^{2} + 3 x + 2) \leq 1

lo g_{2} (2 + 3 x) \leq 4

lo g_{2} (x) + lo g_{2} (x - 1) > lo g_{2} (x + 3)

lo g_{3} (x) > - 9

lo g_{10} (x^{2} - 1) > 0

What is log_{3}(log_{10}(3x))<log_{3}(81) ?
The solution to log_{3}(log_{10}(3x))<log_{3}(81) is 1/3 <x<(10^{81})/3