What is log_{10}(0.9)(3x)>2 ?

The solution to log_{10}(0.9)(3x)>2 is x< 2/(3log_{10)(0.9)}

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log_{10}(0.9)(3x)>2

lo g_{10} (0.9) (3 x) > 2

x < \frac{2}{3 lo g _{10} ( 0.9 )}

Interval Notation

(- \infty, \frac{2}{3 lo g _{10} ( 0.9 )})

Decimal Notation

x < - 14.56956 \dots

Solution steps

lo g_{10} (0.9) (3 x) > 2

Multiply both sides by

- 1

- 3 lo g_{10} (0.9) x < - 2

Divide both sides by

- lo g_{10} (0.9)

3 x < \frac{2}{lo g _{10} ( 0.9 )}

Divide both sides by

3

x < \frac{2}{3 lo g _{10} ( 0.9 )}

5 x + 1 < 6 x + 5 (x + 2) - 3

q, q 13 lo g_{x^{2}} (2 + x) < 1

4 > 11 - 3 x

3 \frac{( x + 1 )}{2} - x - \frac{1}{4} < \frac{1}{2}

\frac{( 4 x + 2 )}{3} \leq \frac{( 25 - 2 x )}{4}

What is log_{10}(0.9)(3x)>2 ?
The solution to log_{10}(0.9)(3x)>2 is x< 2/(3log_{10)(0.9)}