What is log_{6}(14)-1/2 log_{6}(49)+log_{6}(18) ?

The solution to log_{6}(14)-1/2 log_{6}(49)+log_{6}(18) is 2

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simplify log_{6}(14)-1/2 log_{6}(49)+log_{6}(18)

simplify

lo g_{6} (14) - \frac{1}{2} lo g_{6} (49) + lo g_{6} (18)

2

Solution steps

lo g_{6} (14) - \frac{1}{2} lo g_{6} (49) + lo g_{6} (18)

Apply log rule:

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

- \frac{1}{2} lo g_{6} (49) = - lo g_{6} (4 9^{\frac{1}{2}})

= lo g_{6} (14) - lo g_{6} (4 9^{\frac{1}{2}}) + lo g_{6} (18)

Apply log rule:

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

lo g_{6} (14) - lo g_{6} (4 9^{\frac{1}{2}}) = lo g_{6} (\frac{14}{4 9 ^{\frac{1}{2}}})

= lo g_{6} (\frac{14}{4 9 ^{\frac{1}{2}}}) + lo g_{6} (18)

Apply log rule:

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

lo g_{6} (\frac{14}{4 9 ^{\frac{1}{2}}}) + lo g_{6} (18) = lo g_{6} (\frac{14}{4 9 ^{\frac{1}{2}}} \cdot 18)

= lo g_{6} (\frac{14}{4 9 ^{\frac{1}{2}}} \cdot 18)

\frac{14}{4 9 ^{\frac{1}{2}}} \cdot 18 = 36

= lo g_{6} (36)

Factor the number:

36 = 6^{2}

= lo g_{6} (6^{2})

Apply log rule:

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

= 2

ln (\frac{4}{5 x ^{5} y})

lo g_{5} (x) + \frac{1}{3} lo g_{5} (y) - 4 lo g_{5} (z)

lo g_{10} (3 y (5 y + 8))

lo g_{10} (\frac{x}{y ^{6}})

4 lo g_{10} (x) - lo g_{10} (x + 11)

What is log_{6}(14)-1/2 log_{6}(49)+log_{6}(18) ?
The solution to log_{6}(14)-1/2 log_{6}(49)+log_{6}(18) is 2