The solution to x-3>ln(x) is 0 -\W_{-1}(-1/(e^3))

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x-3>ln(x)

Solution

x - 3 > ln (x)

0 < x < - W_{0} (- \frac{1}{e ^{3}}) or x > - W_{- 1} (- \frac{1}{e ^{3}})

Interval Notation

(0, - W_{0} (- \frac{1}{e ^{3}})) \cup (- W_{- 1} (- \frac{1}{e ^{3}}), \infty)

Decimal Notation

0 < x < 0.05246 \dots or x > 4.50524 \dots

Solution steps

x - 3 > ln (x)

Move

ln (x)

to the left side

x - 3 - ln (x) > 0

Move

x

to the right side

- 3 - ln (x) > - x

Move

3

to the right side

- ln (x) > - x + 3

Multiply both sides by

- 1

ln (x) < x - 3

For a>1, if

f (x) < g (x),

then

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

a = e

e^{l n (x)} < e^{x - 3}

Simplify

e^{l n (x)} : x

x < e^{x - 3}

Solve

x < e^{x - 3} : x < - W_{0} (- \frac{1}{e ^{3}}) or x > - W_{- 1} (- \frac{1}{e ^{3}})

Find positive values for logs:

x > 0

Combine the intervals

(x < - W_{0} (- \frac{1}{e ^{3}}) or x > - W_{- 1} (- \frac{1}{e ^{3}})) and x > 0

Merge Overlapping Intervals

0 < x < - W_{0} (- \frac{1}{e ^{3}}) or x > - W_{- 1} (- \frac{1}{e ^{3}})

4 (x + 6) \leq - 12 + 2 x

(x + 2)^{2} > 0

∣ 2 x - 15 ∣ \geq 5

y^{2} + y - 6 = 0

25 a^{2} - 256

What is x-3>ln(x) ?
The solution to x-3>ln(x) is 0<x<-\W_{0}(-1/(e^3))\lor x>-\W_{-1}(-1/(e^3))