What is 5log_{7}(2x)-1/3 log_{7}(5x-1) ?

The solution to 5log_{7}(2x)-1/3 log_{7}(5x-1) is log_{7}((32x^5)/(\sqrt[3]{5x-1)})

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simplify 5log_{7}(2x)-1/3 log_{7}(5x-1)

simplify

5 lo g_{7} (2 x) - \frac{1}{3} lo g_{7} (5 x - 1)

Solution steps

5 lo g_{7} (2 x) - \frac{1}{3} lo g_{7} (5 x - 1)

5 lo g_{7} (2 x) = lo g_{7} ((2 x)^{5})

- \frac{1}{3} lo g_{7} (5 x - 1) = - lo g_{7} ((5 x - 1)^{\frac{1}{3}})

= lo g_{7} ((2 x)^{5}) - lo g_{7} ((5 x - 1)^{\frac{1}{3}})

Apply log rule:

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

lo g_{7} ((2 x)^{5}) - lo g_{7} ((5 x - 1)^{\frac{1}{3}}) = lo g_{7} (\frac{( 2 x ) ^{5}}{( 5 x - 1 ) ^{\frac{1}{3}}})

= lo g_{7} (\frac{( 2 x ) ^{5}}{( 5 x - 1 ) ^{\frac{1}{3}}})

Simplify

(2 x)^{5} : 32 x^{5}

= lo g_{7} (\frac{32 x ^{5}}{( 5 x - 1 ) ^{\frac{1}{3}}})

Apply exponent rule:

ln (θ e^{θ} x^{- θ - 1})

lo g_{10} (x^{3} \cdot y^{2})

lo g_{2} ((3 x)^{3})

lo g_{6} (7 x^{2})

lo g_{4} (\frac{x ^{2} - 2 x - 3}{x ^{2} - 3 x})

What is 5log_{7}(2x)-1/3 log_{7}(5x-1) ?
The solution to 5log_{7}(2x)-1/3 log_{7}(5x-1) is log_{7}((32x^5)/(\sqrt[3]{5x-1)})