What is ln(x)-ln(x-2)<5 ?

The solution to ln(x)-ln(x-2) (2e^5)/(e^5-1)

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ln(x)-ln(x-2)<5

ln (x) - ln (x - 2) < 5

x > \frac{2 e ^{5}}{e ^{5} - 1}

Interval Notation

(\frac{2 e ^{5}}{e ^{5} - 1}, \infty)

Decimal Notation

x > 2.01356 \dots

Solution steps

ln (x) - ln (x - 2) < 5

Apply log rules

\frac{x}{x - 2} < e^{5} and \frac{x}{x - 2} > 0

\frac{x}{x - 2} < e^{5} and \frac{x}{x - 2} > 0 : x < 0 or x > \frac{2 e ^{5}}{e ^{5} - 1}

x < 0 or x > \frac{2 e ^{5}}{e ^{5} - 1}

Find positive values for logs:

x > 2

Combine the intervals

(x < 0 or x > \frac{2 e ^{5}}{e ^{5} - 1}) and x > 2

Merge Overlapping Intervals

x > \frac{2 e ^{5}}{e ^{5} - 1}

x - 3 \leq 4 x + 9

\frac{6 ^{0} t u ^{- 5}}{2 ^{5}}

3 \cdot x^{2} + x = 0

- 16 t^{2} + 70 t - 25 = 0

- \frac{7}{8} + (- \frac{1}{2})