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Popular >

4^x<2^{x+1}+3

Solution

4^{x} < 2^{x + 1} + 3

Solution

x < lo g_{2} (3)

+2

Interval Notation

(- \infty, lo g_{2} (3))

Decimal Notation

x < 1.58496 \dots

Solution steps

4^{x} < 2^{x + 1} + 3

Apply exponent rules

(2^{x})^{2} < 2^{x} \cdot 2 + 3

Let

u = 2^{x}

u^{2} < u \cdot 2 + 3

u^{2} < u \cdot 2 + 3 : - 1 < u < 3

- 1 < u < 3

Substitute back

u = 2^{x}

- 1 < 2^{x} < 3

If

a < u < b

then

a < u and u < b

- 1 < 2^{x} and 2^{x} < 3

- 1 < 2^{x} :

True for all

x

2^{x} < 3 : x < lo g_{2} (3)

Combine the intervals

True for all x and x < lo g_{2} (3)

Merge Overlapping Intervals

x < lo g_{2} (3)

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3^{x} + 3^{2 - x} > 10

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\frac{[ 2 ^{1 - x} - 2 ^{x} + 1 ]}{[ 2 ^{x} - 1 ]} < 0

e^{x - 1} > 0

e^{- x} < 1

(2/3)^{-x^2-2x+12}>= (9/4)^{((x-6))/2}

(\frac{2}{3})^{- x^{2} - 2 x + 12} \geq (\frac{9}{4})^{\frac{( x - 6 )}{2}}

Frequently Asked Questions (FAQ)

What is 4^x<2^{x+1}+3 ?
The solution to 4^x<2^{x+1}+3 is x<log_{2}(3)