What is 2log_{10}(x)+log_{10}(6)-log_{10}(30) ?

The solution to 2log_{10}(x)+log_{10}(6)-log_{10}(30) is log_{10}((x^2)/5)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

simplify 2log_{10}(x)+log_{10}(6)-log_{10}(30)

simplify

2 lo g_{10} (x) + lo g_{10} (6) - lo g_{10} (30)

lo g_{10} (\frac{x ^{2}}{5})

Solution steps

2 lo g_{10} (x) + lo g_{10} (6) - lo g_{10} (30)

Apply log rule:

b \cdot lo g_{a} (x) = lo g_{a} (x^{b})

2 lo g_{10} (x) = lo g_{10} (x^{2})

= lo g_{10} (x^{2}) + lo g_{10} (6) - lo g_{10} (30)

Apply log rule:

lo g_{a} (x) + lo g_{a} (y) = lo g_{a} (x y)

lo g_{10} (x^{2}) + lo g_{10} (6) = lo g_{10} (x^{2} \cdot 6)

= lo g_{10} (x^{2} \cdot 6) - lo g_{10} (30)

Apply log rule:

lo g_{a} (x) - lo g_{a} (y) = lo g_{a} (\frac{x}{y})

lo g_{10} (x^{2} \cdot 6) - lo g_{10} (30) = lo g_{10} (\frac{x ^{2} \cdot 6}{30})

= lo g_{10} (\frac{x ^{2} \cdot 6}{30})

Cancel

\frac{x ^{2} \cdot 6}{30} : \frac{x ^{2}}{5}

= lo g_{10} (\frac{x ^{2}}{5})

ln (x) + ln (- ln (x) - ln (4))

lo g_{2} (\frac{s ^{5}}{7 t ^{2}})

lo g_{2} (x + 1) - lo g_{2} (x)

4 [lo g_{2} (k) - lo g_{2} (k - t)]

ln (28) 0.125 + ln (\frac{3}{5}) 0.6 - 49 ln (7 e^{3})

What is 2log_{10}(x)+log_{10}(6)-log_{10}(30) ?
The solution to 2log_{10}(x)+log_{10}(6)-log_{10}(30) is log_{10}((x^2)/5)