What is log_{3}(7x-6)<log_{3}(4x+9) ?

The solution to log_{3}(7x-6)<log_{3}(4x+9) is 6/7 <x<5

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log_{3}(7x-6)<log_{3}(4x+9)

lo g_{3} (7 x - 6) < lo g_{3} (4 x + 9)

\frac{6}{7} < x < 5

Interval Notation

(\frac{6}{7}, 5)

Decimal Notation

0.85714 \dots < x < 5

Solution steps

lo g_{3} (7 x - 6) < lo g_{3} (4 x + 9)

Apply log rules

7 x - 6 < 4 x + 9 and 7 x - 6 > 0

7 x - 6 < 4 x + 9 and 7 x - 6 > 0 : \frac{6}{7} < x < 5

\frac{6}{7} < x < 5

(\frac{1}{2})^{2} + (\frac{1}{2})^{2}

\frac{3 - \frac{1}{3} x}{3 + \frac{1}{2}} \geq \frac{3 x - \frac{5}{2}}{1 - \frac{2}{3}}

- 2 (5 + 6 c) + 16 = - 90

(2 u^{2} v^{3})^{3} (3 u^{- 3} v)^{2}

\frac{2}{2} + \frac{1 - 2}{- 1}

What is log_{3}(7x-6)<log_{3}(4x+9) ?
The solution to log_{3}(7x-6)<log_{3}(4x+9) is 6/7 <x<5