What is 2/(2^{x-1)}<= 0.05 ?

The solution to 2/(2^{x-1)} = log_{2}(40)+1

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2/(2^{x-1)}<= 0.05

\frac{2}{2 ^{x - 1}} \leq 0.05

x \geq lo g_{2} (40) + 1

Interval Notation

[lo g_{2} (40) + 1, \infty)

Decimal Notation

x \geq 6.32192 \dots

Solution steps

\frac{2}{2 ^{x - 1}} \leq 0.05

Divide both sides by

2

\frac{\frac{2}{2 ^{x - 1}}}{2} \leq \frac{0.05}{2}

Simplify

\frac{1}{2 ^{x - 1}} \leq \frac{0.05}{2}

Rewrite in standard form

\frac{2 - 0.05 \cdot 2 ^{x - 1}}{2 ^{x - 1}} \leq 0

Identify the intervals

2^{x - 1} < 0 or 2^{x - 1} \geq 40

2^{x - 1} < 0 :

No Solution

2^{x - 1} \geq 40 : x \geq lo g_{2} (40) + 1

Combine the intervals

No Solution or x \geq lo g_{2} (40) + 1

Merge Overlapping Intervals

x \geq lo g_{2} (40) + 1

\frac{20 π}{180}

(4 - x)^{2} (x - \frac{9}{2}) < 0

- \frac{2}{3} (- 2)

x^{2} + 8^{2} = 1 5^{2}

- 120

What is 2/(2^{x-1)}<= 0.05 ?
The solution to 2/(2^{x-1)}<= 0.05 is x>= log_{2}(40)+1