What is 2log_{10}(4x)-log_{10}(8) ?

The solution to 2log_{10}(4x)-log_{10}(8) is log_{10}(2x^2)

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simplify 2log_{10}(4x)-log_{10}(8)

simplify

2 lo g_{10} (4 x) - lo g_{10} (8)

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

Solution steps

2 lo g_{10} (4 x) - lo g_{10} (8)

Apply log rule:

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

2 lo g_{10} (4 x) = lo g_{10} ((4 x)^{2})

= lo g_{10} ((4 x)^{2}) - lo g_{10} (8)

Apply log rule:

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

lo g_{10} ((4 x)^{2}) - lo g_{10} (8) = lo g_{10} (\frac{( 4 x ) ^{2}}{8})

= lo g_{10} (\frac{( 4 x ) ^{2}}{8})

Simplify

\frac{( 4 x ) ^{2}}{8} : 2 x^{2}

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

lo g_{10} (4 x - 1) - lo g_{10} (x)

lo g_{10} (p^{n} (1 - p)^{n (x - 1)})

(\frac{ln ( 2 )}{2} t) - ln (- 7 t)

lo g_{5} (3) + lo g_{5} (4)

What is 2log_{10}(4x)-log_{10}(8) ?
The solution to 2log_{10}(4x)-log_{10}(8) is log_{10}(2x^2)