What is -x+ln(1+x)<ln(0.01) ?

The solution to -x+ln(1+x) -\W_{-1}(-1/(100e))-1

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-x+ln(1+x)<ln(0.01)

- x + ln (1 + x) < ln (0.01)

- 1 < x < - W_{0} (- \frac{1}{100 e}) - 1 or x > - W_{- 1} (- \frac{1}{100 e}) - 1

Interval Notation

(- 1, - W_{0} (- \frac{1}{100 e}) - 1) \cup (- W_{- 1} (- \frac{1}{100 e}) - 1, \infty)

Decimal Notation

- 1 < x < - 0.99630 \dots or x > 6.63835 \dots

Solution steps

- x + ln (1 + x) < ln (0.01)

Move

x

to the right side

ln (1 + x) < ln (0.01) + x

For a>1, if

f (x) < g (x),

then

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

a = e

e^{l n (1 + x)} < e^{l n (0.01) + x}

Simplify

e^{l n (1 + x)} : 1 + x

Simplify

e^{l n (0.01) + x} : 0.01 e^{x}

1 + x < 0.01 e^{x}

Solve

1 + x < 0.01 e^{x} : x < - W_{0} (- \frac{1}{100 e}) - 1 or x > - W_{- 1} (- \frac{1}{100 e}) - 1

Find positive values for logs:

x > - 1

Combine the intervals

(x < - W_{0} (- \frac{1}{100 e}) - 1 or x > - W_{- 1} (- \frac{1}{100 e}) - 1) and x > - 1

Merge Overlapping Intervals

- 1 < x < - W_{0} (- \frac{1}{100 e}) - 1 or x > - W_{- 1} (- \frac{1}{100 e}) - 1

3 (x + 4)^{2} = 21

g^{2} - 14 g + 13

\frac{2}{3} x - \frac{5}{3} \leq 3 x + \frac{1}{3} \leq 4 x - \frac{7}{3}

- 8.31 \geq - 2 - f

\frac{16 x}{4 x ^{2} y}

What is -x+ln(1+x)<ln(0.01) ?
The solution to -x+ln(1+x)<ln(0.01) is -1<x<-\W_{0}(-1/(100e))-1\lor x>-\W_{-1}(-1/(100e))-1