What is ln(5x)+ln(3x)>9 ?

The solution to ln(5x)+ln(3x)>9 is x>(e^4sqrt(e))/(sqrt(15))

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ln(5x)+ln(3x)>9

ln (5 x) + ln (3 x) > 9

x > \frac{e ^{4} e}{15}

Interval Notation

(\frac{e ^{4} e}{15}, \infty)

Decimal Notation

x > 23.24232 \dots

Solution steps

ln (5 x) + ln (3 x) > 9

Apply log rules

15 x^{2} > e^{9}

15 x^{2} > e^{9} : x < - \frac{e ^{4} e}{15} or x > \frac{e ^{4} e}{15}

x < - \frac{e ^{4} e}{15} or x > \frac{e ^{4} e}{15}

Find positive values for logs:

x > 0

Combine the intervals

(x < - \frac{e ^{4} e}{15} or x > \frac{e ^{4} e}{15}) and x > 0

Merge Overlapping Intervals

x > \frac{e ^{4} e}{15}

2 x^{2} + 6 x = 1

2 x^{2} + x - 6 \geq 0

(\frac{4}{9})^{2}

\frac{x ^{2}}{18} - \frac{x}{9} = 1

3 x + 12 = \frac{x + 7}{3}

What is ln(5x)+ln(3x)>9 ?
The solution to ln(5x)+ln(3x)>9 is x>(e^4sqrt(e))/(sqrt(15))