What is log_{10}(8)(k^2-64)-log_{10}(8k)-8 ?

The solution to log_{10}(8)(k^2-64)-log_{10}(8k)-8 is log_{10}((8^{k^2-65})/k)-8

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simplify log_{10}(8)(k^2-64)-log_{10}(8k)-8

simplify

lo g_{10} (8) (k^{2} - 64) - lo g_{10} (8 k) - 8

lo g_{10} (\frac{8 ^{k^{2} - 65}}{k}) - 8

Solution steps

lo g_{10} (8) (k^{2} - 64) - lo g_{10} (8 k) - 8

Apply log rule:

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

lo g_{10} (8) (k^{2} - 64) = lo g_{10} (8^{k^{2} - 64})

= lo g_{10} (8^{k^{2} - 64}) - lo g_{10} (8 k) - 8

Simplify

lo g_{10} (8^{k^{2} - 64}) - lo g_{10} (8 k) : lo g_{10} (\frac{8 ^{k^{2} - 65}}{k})

= lo g_{10} (\frac{8 ^{k^{2} - 65}}{k}) - 8

ln ((x + 1)^{5} 3 x)

lo g_{w} (\frac{7 x}{5})

ln (a^{10})

lo g_{10} (12) - lo g_{10} (2)

\frac{1}{4} ln (17)

What is log_{10}(8)(k^2-64)-log_{10}(8k)-8 ?
The solution to log_{10}(8)(k^2-64)-log_{10}(8k)-8 is log_{10}((8^{k^2-65})/k)-8