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}} \leq \frac{1}{81}

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

Interval Notation

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

Decimal Notation

x \leq - 3.63092 \dots

Solution steps

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

Divide both sides by

2

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

Simplify

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

Rewrite in standard form

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

Identify the intervals

3^{1 - x} < 0 or 3^{1 - x} \geq 162

3^{1 - x} < 0 :

No Solution

3^{1 - x} \geq 162 : x \leq - lo g_{3} (162) + 1

Combine the intervals

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

Merge Overlapping Intervals

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

3 y + 5 y = 64

lo g_{5} (x) \leq lo g_{x} (5)

8 - 22

3 x^{2} - 6 x - 10 = 0

(10 - 5) (10 + 2)

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