What is ln(3-x)<= 1+ln(x) ?

The solution to ln(3-x)<= 1+ln(x) is 3/(1+e)<= x<3

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ln(3-x)<= 1+ln(x)

ln (3 - x) \leq 1 + ln (x)

\frac{3}{1 + e} \leq x < 3

Interval Notation

[\frac{3}{1 + e}, 3)

Decimal Notation

0.80682 \dots \leq x < 3

Solution steps

ln (3 - x) \leq 1 + ln (x)

Apply log rules

3 - x \leq e x and 3 - x > 0

3 - x \leq e x and 3 - x > 0 : \frac{3}{1 + e} \leq x < 3

\frac{3}{1 + e} \leq x < 3

Find positive values for logs:

0 < x < 3

Combine the intervals

\frac{3}{1 + e} \leq x < 3 and 0 < x < 3

Merge Overlapping Intervals

\frac{3}{1 + e} \leq x < 3

x > x

t^{2} - 4 t + 5 = 0

\frac{x + 1}{1 - x} \geq 0

y (y + 1) - 4 = 0

2 (2 q + 1.5) = 18 - q