What is log_{10}(25x)-log_{10}(x-2) ?

The solution to log_{10}(25x)-log_{10}(x-2) is log_{10}((25x)/(x-2))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

simplify log_{10}(25x)-log_{10}(x-2)

simplify

lo g_{10} (25 x) - lo g_{10} (x - 2)

lo g_{10} (\frac{25 x}{x - 2})

Solution steps

lo g_{10} (25 x) - lo g_{10} (x - 2)

Simplify

lo g_{10} (25 x) : 2 lo g_{10} (5) + lo g_{10} (x)

= (2 lo g_{10} (5) + lo g_{10} (x)) - lo g_{10} (x - 2)

Apply rule:

(a) = a

(2 lo g_{10} (5) + lo g_{10} (x)) = 2 lo g_{10} (5) + lo g_{10} (x)

= 2 lo g_{10} (5) + lo g_{10} (x) - lo g_{10} (x - 2)

Apply log rule:

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

2 lo g_{10} (5) = lo g_{10} (5^{2})

= lo g_{10} (5^{2}) + lo g_{10} (x) - lo g_{10} (x - 2)

Apply log rule:

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

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

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

Apply log rule:

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

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

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

Simplify

\frac{5 ^{2} x}{x - 2} : \frac{25 x}{x - 2}

= lo g_{10} (\frac{25 x}{x - 2})

10 lo g_{10} (\frac{5 x 1 0 ^{- 8}}{1 0 ^{- 12}})

lo g_{5} (\frac{x ^{e}}{y π})

ln (2 e^{2 x} x)

lo g_{2} (8 x)

lo g_{3} (54) + lo g_{3} (10) - lo g_{3} (54)

What is log_{10}(25x)-log_{10}(x-2) ?
The solution to log_{10}(25x)-log_{10}(x-2) is log_{10}((25x)/(x-2))