What is 2log_{10}(b)-log_{10}(k) ?

The solution to 2log_{10}(b)-log_{10}(k) is log_{10}((b^2)/k)

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simplify 2log_{10}(b)-log_{10}(k)

simplify

2 lo g_{10} (b) - lo g_{10} (k)

lo g_{10} (\frac{b ^{2}}{k})

Solution steps

2 lo g_{10} (b) - lo g_{10} (k)

Apply log rule:

a lo g_{c} (b) = lo g_{c} (b^{a})

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

= lo g_{10} (b^{2}) - lo g_{10} (k)

Apply log rule:

lo g_{c} (a) - lo g_{c} (b) = lo g_{c} (\frac{a}{b})

lo g_{10} (b^{2}) - lo g_{10} (k) = lo g_{10} (\frac{b ^{2}}{k})

= lo g_{10} (\frac{b ^{2}}{k})

lo g_{5} (4 \cdot 7) + lo g_{5} (2)

3 lo g_{10} (x^{5}) - 2 lo g_{10} (x^{4})

lo g_{10} (\frac{x ^{2} \cdot y}{3 z})

lo g_{2} (\frac{15 x ( x - 1 )}{( 3 x + 4 ) ( 2 - x )})

What is 2log_{10}(b)-log_{10}(k) ?
The solution to 2log_{10}(b)-log_{10}(k) is log_{10}((b^2)/k)