What is 4^x<2^x+1+3 ?

The solution to 4^x<2^x+1+3 is x<log_{2}(sqrt(17)+1)-1

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4^x<2^x+1+3

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

x < lo g_{2} (17 + 1) - 1

Interval Notation

(- \infty, lo g_{2} (17 + 1) - 1)

Decimal Notation

x < 1.35701 \dots

Solution steps

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

Apply exponent rules

(2^{x})^{2} < 2^{x} + 1 + 3

Let

u = 2^{x}

u^{2} < u + 1 + 3

u^{2} < u + 1 + 3 : \frac{- 17 + 1}{2} < u < \frac{17 + 1}{2}

\frac{- 17 + 1}{2} < u < \frac{17 + 1}{2}

Substitute back

u = 2^{x}

\frac{- 17 + 1}{2} < 2^{x} < \frac{17 + 1}{2}

a < u < b

then

a < u and u < b

\frac{- 17 + 1}{2} < 2^{x} and 2^{x} < \frac{17 + 1}{2}

\frac{- 17 + 1}{2} < 2^{x} :

True for all

x

2^{x} < \frac{17 + 1}{2} : x < lo g_{2} (17 + 1) - 1

Combine the intervals

True for all x and x < lo g_{2} (17 + 1) - 1

Merge Overlapping Intervals

x < lo g_{2} (17 + 1) - 1

5^{1 - x} < 5^{4 - 5 x}

x, (\frac{3}{2})^{1 + r \cdot l^{2}} > (\frac{2}{3})^{3 x - 1}

0. 3^{7} > 0. 3^{x^{2} + 6 x}

5^{3 - x^{2}} \leq 5^{2}

e^{x} - 1 > 0