What is 3x+1<2*log_{2}(x+4) ?

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

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3x+1<2*log_{2}(x+4)

3 x + 1 < 2 \cdot lo g_{2} (x + 4)

- 4 < x < \frac{- 2 W _{0} ( - \frac{3 l n ( 2 )}{2 e ^{\frac{11 l n ( 2 )}{2}}} ) - 12 ln ( 2 )}{3 ln ( 2 )} or x > \frac{- 2 W _{- 1} ( - \frac{3 l n ( 2 )}{2 e ^{\frac{11 l n ( 2 )}{2}}} ) - 12 ln ( 2 )}{3 ln ( 2 )}

Interval Notation

- 4, \frac{- 2 W _{0} ( - \frac{3 l n ( 2 )}{2 e ^{\frac{11 l n ( 2 )}{2}}} ) - 12 ln ( 2 )}{3 ln ( 2 )} \cup \frac{- 2 W _{- 1} ( - \frac{3 l n ( 2 )}{2 e ^{\frac{11 l n ( 2 )}{2}}} ) - 12 ln ( 2 )}{3 ln ( 2 )}, \infty

Decimal Notation

- 4 < x < - 3.97737 \dots or x > 1.26413 \dots

Solution steps

3 x + 1 < 2 lo g_{2} (x + 4)

Move

2 lo g_{2} (x + 4)

to the left side

3 x + 1 - 2 lo g_{2} (x + 4) < 0

Move

3 x

to the right side

1 - 2 lo g_{2} (x + 4) < - 3 x

Move

1

to the right side

- 2 lo g_{2} (x + 4) < - 3 x - 1

Multiply both sides by

- 1

2 lo g_{2} (x + 4) > 3 x + 1

Divide both sides by

2

lo g_{2} (x + 4) > \frac{3 x + 1}{2}

For a>1, if

f (x) > g (x),

then

a^{f (x)} > a^{g (x)}

a = 2

2^{l o g_{2} (x + 4)} > 2^{\frac{3 x + 1}{2}}

Simplify

2^{l o g_{2} (x + 4)} : x + 4

x + 4 > 2^{\frac{3 x + 1}{2}}

Solve

x + 4 > 2^{\frac{3 x + 1}{2}} : x < \frac{- 2 W _{0} ( - \frac{3 l n ( 2 )}{2 e ^{\frac{11 l n ( 2 )}{2}}} ) - 12 ln ( 2 )}{3 ln ( 2 )} or x > \frac{- 2 W _{- 1} ( - \frac{3 l n ( 2 )}{2 e ^{\frac{11 l n ( 2 )}{2}}} ) - 12 ln ( 2 )}{3 ln ( 2 )}

Find positive values for logs:

x > - 4

Combine the intervals

x < \frac{- 2 W _{0} ( - \frac{3 l n ( 2 )}{2 e ^{\frac{11 l n ( 2 )}{2}}} ) - 12 ln ( 2 )}{3 ln ( 2 )} or x > \frac{- 2 W _{- 1} ( - \frac{3 l n ( 2 )}{2 e ^{\frac{11 l n ( 2 )}{2}}} ) - 12 ln ( 2 )}{3 ln ( 2 )} and x > - 4

Merge Overlapping Intervals

- 4 < x < \frac{- 2 W _{0} ( - \frac{3 l n ( 2 )}{2 e ^{\frac{11 l n ( 2 )}{2}}} ) - 12 ln ( 2 )}{3 ln ( 2 )} or x > \frac{- 2 W _{- 1} ( - \frac{3 l n ( 2 )}{2 e ^{\frac{11 l n ( 2 )}{2}}} ) - 12 ln ( 2 )}{3 ln ( 2 )}

lo g_{10} (x - 2) (3 x^{2} - x - 1) > 0

2 - lo g_{10} (4 x) \geq 0

lo g_{2} (x - 1) > - 1

lo g_{x - 2} (3 x^{2} - x - 1) > 0

lo g_{4} (x^{2} + 2 x) > lo g_{2} (x - 4)