What is 4^{x-1}+2^{x+1}>= 2^5 ?

The solution to 4^{x-1}+2^{x+1}>= 2^5 is x>= 3

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4^{x-1}+2^{x+1}>= 2^5

4^{x - 1} + 2^{x + 1} \geq 2^{5}

x \geq 3

Interval Notation

[3, \infty)

Solution steps

4^{x - 1} + 2^{x + 1} \geq 2^{5}

Apply exponent rules

\frac{( 2 ^{x} ) ^{2}}{4} + 2^{x} \cdot 2 \geq 2^{5}

Let

u = 2^{x}

\frac{u ^{2}}{4} + u \cdot 2 \geq 2^{5}

\frac{u ^{2}}{4} + u \cdot 2 \geq 2^{5} : u \leq - 16 or u \geq 8

u \leq - 16 or u \geq 8

Substitute back

u = 2^{x}

2^{x} \leq - 16 or 2^{x} \geq 8

2^{x} \leq - 16 :

No Solution

2^{x} \geq 8 : x \geq 3

Combine the intervals

No Solution or x \geq 3

Merge Overlapping Intervals

x \geq 3

2^{x + 2} > (\frac{1}{4})^{\frac{1}{x}}

e^{- 2 x^{2}} - 1 > 0

3^{x} \leq 4^{x}

9^{x} + 3^{x} - 6 > 0

\frac{5 ^{x} + 625}{( 5 ^{x} - 130 )} \leq 0