What is log_{b}(5)+3log_{b}(10)-log_{b}(40) ?

The solution to log_{b}(5)+3log_{b}(10)-log_{b}(40) is 3log_{b}(5)

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simplify log_{b}(5)+3log_{b}(10)-log_{b}(40)

simplify

lo g_{b} (5) + 3 lo g_{b} (10) - lo g_{b} (40)

3 lo g_{b} (5)

Solution steps

lo g_{b} (5) + 3 lo g_{b} (10) - lo g_{b} (40)

Apply log rule:

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

3 lo g_{b} (10) = lo g_{b} (1 0^{3})

= lo g_{b} (5) + lo g_{b} (1 0^{3}) - lo g_{b} (40)

Apply log rule:

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

lo g_{b} (5) + lo g_{b} (1 0^{3}) = lo g_{b} (5 \cdot 1 0^{3})

= lo g_{b} (5 \cdot 1 0^{3}) - lo g_{b} (40)

5 \cdot 1 0^{3} = 5000

= lo g_{b} (5000) - lo g_{b} (40)

Apply log rule:

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

lo g_{b} (5000) - lo g_{b} (40) = lo g_{b} (\frac{5000}{40})

= lo g_{b} (\frac{5000}{40})

Divide the numbers:

\frac{5000}{40} = 125

= lo g_{b} (125)

Factor the number:

125 = 5^{3}

= lo g_{b} (5^{3})

Apply log rule:

lo g_{a} (x^{b}) = b \cdot lo g_{a} (x), x > 0

lo g_{b} (5^{3}) = 3 lo g_{b} (5)

= 3 lo g_{b} (5)

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

lo g_{4} (13 b^{8} c)

lo g_{9} (\frac{9}{x})

ln (x) - 8 ln (x + 5)

lo g_{5} (\frac{125}{x})

What is log_{b}(5)+3log_{b}(10)-log_{b}(40) ?
The solution to log_{b}(5)+3log_{b}(10)-log_{b}(40) is 3log_{b}(5)