What is log_{4}(3)+1/2 log_{4}(8)-log_{4}(2) ?

The solution to log_{4}(3)+1/2 log_{4}(8)-log_{4}(2) is log_{4}(3sqrt(2))

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simplify log_{4}(3)+1/2 log_{4}(8)-log_{4}(2)

simplify

lo g_{4} (3) + \frac{1}{2} lo g_{4} (8) - lo g_{4} (2)

lo g_{4} (32)

Decimal Notation

1.04248 \dots

Solution steps

lo g_{4} (3) + \frac{1}{2} lo g_{4} (8) - lo g_{4} (2)

Apply log rule:

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

\frac{1}{2} lo g_{4} (8) = lo g_{4} (8^{\frac{1}{2}})

= lo g_{4} (3) + lo g_{4} (8^{\frac{1}{2}}) - lo g_{4} (2)

Apply log rule:

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

lo g_{4} (3) + lo g_{4} (8^{\frac{1}{2}}) = lo g_{4} (3 \cdot 8^{\frac{1}{2}})

= lo g_{4} (3 \cdot 8^{\frac{1}{2}}) - lo g_{4} (2)

Apply log rule:

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

lo g_{4} (3 \cdot 8^{\frac{1}{2}}) - lo g_{4} (2) = lo g_{4} (\frac{3 \cdot 8 ^{\frac{1}{2}}}{2})

= lo g_{4} (\frac{3 \cdot 8 ^{\frac{1}{2}}}{2})

Cancel

\frac{3 \cdot 8 ^{\frac{1}{2}}}{2} : 3 \cdot 2^{\frac{1}{2}}

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

Apply exponent rule:

a^{\frac{1}{2}} = a

2^{\frac{1}{2}} = 2

= lo g_{4} (32)

ln (4) - ln (6)

ln (4) - ln (5)

2 lo g_{5} (x) + 12 lo g_{5} (y)

3 ln (x) + 6 ln (y) - 2 ln (z)

What is log_{4}(3)+1/2 log_{4}(8)-log_{4}(2) ?
The solution to log_{4}(3)+1/2 log_{4}(8)-log_{4}(2) is log_{4}(3sqrt(2))