What is log_{10}(2a)+log_{10}(b)-2log_{10}(ab) ?

The solution to log_{10}(2a)+log_{10}(b)-2log_{10}(ab) is log_{10}(2/(ab))

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simplify log_{10}(2a)+log_{10}(b)-2log_{10}(ab)

simplify

lo g_{10} (2 a) + lo g_{10} (b) - 2 lo g_{10} (ab)

lo g_{10} (\frac{2}{ab})

Solution steps

lo g_{10} (2 a) + lo g_{10} (b) - 2 lo g_{10} (ab)

Apply log rule:

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

- 2 lo g_{10} (ab) = - lo g_{10} ((ab)^{2})

= lo g_{10} (2 a) + lo g_{10} (b) - lo g_{10} ((ab)^{2})

Apply log rule:

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

lo g_{10} (2 a) + lo g_{10} (b) = lo g_{10} (2 ab)

= lo g_{10} (2 ab) - lo g_{10} ((ab)^{2})

Apply log rule:

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

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

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

Cancel

\frac{2 ab}{( ab ) ^{2}} : \frac{2}{ab}

= lo g_{10} (\frac{2}{ab})

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

lo g_{10} (\frac{x ^{3}}{2 x + 4})

lo g_{5} (x) + 2 lo g_{5} (y) - \frac{1}{2} lo g_{5} (17)

lo g_{10} (a b^{x})

lo g_{10} (\frac{x ( x + 2 )}{( x + 4 ) ^{2}})

What is log_{10}(2a)+log_{10}(b)-2log_{10}(ab) ?
The solution to log_{10}(2a)+log_{10}(b)-2log_{10}(ab) is log_{10}(2/(ab))