What is 2/(3^{1-x)}< 1/81 ?

The solution to 2/(3^{1-x)}< 1/81 is x<-log_{3}(162)+1

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2/(3^{1-x)}< 1/81

\frac{2}{3 ^{1 - x}} < \frac{1}{81}

x < - lo g_{3} (162) + 1

Interval Notation

(- \infty, - lo g_{3} (162) + 1)

Decimal Notation

x < - 3.63092 \dots

Solution steps

\frac{2}{3 ^{1 - x}} < \frac{1}{81}

Divide both sides by

2

\frac{\frac{2}{3 ^{1 - x}}}{2} < \frac{\frac{1}{81}}{2}

Simplify

\frac{1}{3 ^{1 - x}} < \frac{1}{162}

Rewrite in standard form

\frac{162 - 3 ^{1 - x}}{3 ^{1 - x}} < 0

Identify the intervals

3^{1 - x} < 0 or 3^{1 - x} > 162

3^{1 - x} < 0 :

No Solution

3^{1 - x} > 162 : x < - lo g_{3} (162) + 1

Combine the intervals

No Solution or x < - lo g_{3} (162) + 1

Merge Overlapping Intervals

x < - lo g_{3} (162) + 1

x^{4} - 11 x^{2} + 24

x^{2} - 12 x + 10 = 0

x^{2} + 3 x + 8 \leq 0

x^{2} + x + 9 = 0

25 x^{2} - 60 x + 36

What is 2/(3^{1-x)}< 1/81 ?
The solution to 2/(3^{1-x)}< 1/81 is x<-log_{3}(162)+1