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Popular >

(2^n)/(2^n+2-3)< 1/4

Solution

\frac{2 ^{n}}{2 ^{n} + 2 - 3} < \frac{1}{4}

Solution

n < 0

+1

Interval Notation

(- \infty, 0)

Solution steps

\frac{2 ^{n}}{2 ^{n} + 2 - 3} < \frac{1}{4}

Let

u = 2^{n}

\frac{u}{u + 2 - 3} < \frac{1}{4}

\frac{u}{u + 2 - 3} < \frac{1}{4} : - \frac{1}{3} < u < 1

- \frac{1}{3} < u < 1

Substitute back

u = 2^{n}

- \frac{1}{3} < 2^{n} < 1

If

a < u < b

then

a < u and u < b

- \frac{1}{3} < 2^{n} and 2^{n} < 1

- \frac{1}{3} < 2^{n} :

True for all

n

2^{n} < 1 : n < 0

Combine the intervals

True for all n and n < 0

Merge Overlapping Intervals

n < 0

Popular Examples

4^{1+x}+4^x>= 5

4^{1 + x} + 4^{x} \geq 5

e^{x} \geq 2

e^{x} \geq 1

3^{x} < 4^{x}

([2^{x-1}-1])/([2^{x+1)+1]}<2

\frac{[ 2 ^{x - 1} - 1 ]}{[ 2 ^{x + 1} + 1 ]} < 2

Frequently Asked Questions (FAQ)

What is (2^n)/(2^n+2-3)< 1/4 ?
The solution to (2^n)/(2^n+2-3)< 1/4 is n<0