What is log_{10}((1/2)^4)+log_{10}((2/3)^4) ?

The solution to log_{10}((1/2)^4)+log_{10}((2/3)^4) is -4log_{10}(3)

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simplify log_{10}((1/2)^4)+log_{10}((2/3)^4)

simplify

lo g_{10} ((\frac{1}{2})^{4}) + lo g_{10} ((\frac{2}{3})^{4})

- 4 lo g_{10} (3)

Decimal Notation

- 1.90848 \dots

Solution steps

lo g_{10} ((\frac{1}{2})^{4}) + lo g_{10} ((\frac{2}{3})^{4})

Apply log rule:

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

lo g_{10} ((\frac{1}{2})^{4}) + lo g_{10} ((\frac{2}{3})^{4}) = lo g_{10} ((\frac{1}{2})^{4} (\frac{2}{3})^{4})

= lo g_{10} ((\frac{1}{2})^{4} (\frac{2}{3})^{4})

Simplify

(\frac{1}{2})^{4} (\frac{2}{3})^{4} : \frac{1}{81}

= lo g_{10} (\frac{1}{81})

\frac{1}{81} = 8 1^{- 1}

= lo g_{10} (8 1^{- 1})

Apply log rule:

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

lo g_{10} (8 1^{- 1}) = - 1 \cdot lo g_{10} (81)

= - 1 \cdot lo g_{10} (81)

Apply rule:

1 \cdot a = a

= - lo g_{10} (81)

Factor the number:

81 = 3^{4}

= - lo g_{10} (3^{4})

Apply log rule:

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

- lo g_{10} (3^{4}) = - 4 lo g_{10} (3)

= - 4 lo g_{10} (3)

11 ln (x - 2) - 3 ln (x)

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4 lo g_{10} (9) - 4 lo g_{10} (x)

What is log_{10}((1/2)^4)+log_{10}((2/3)^4) ?
The solution to log_{10}((1/2)^4)+log_{10}((2/3)^4) is -4log_{10}(3)