What is the answer to log_{10}(\sqrt[5]{x^2-21x})= 2/5 ?

The answer to log_{10}(\sqrt[5]{x^2-21x})= 2/5 is x=25,x=-4

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log_{10}(\sqrt[5]{x^2-21x})= 2/5

lo g_{10} (5 x^{2} - 21 x) = \frac{2}{5}

x = 25, x = - 4

Solution steps

lo g_{10} (5 x^{2} - 21 x) = \frac{2}{5}

Apply log rules

5 x^{2} - 21 x = 1 0^{\frac{2}{5}}

Solve

5 x^{2} - 21 x = 1 0^{\frac{2}{5}} : x = 25, x = - 4

x = 25, x = - 4

2 lo g_{10} (x) - lo g_{10} (x + 6) = 0

10 lo g_{5} (9 x - 5) = 35

2 lo g_{10} (x) = 3 + lo g_{10} (\frac{x}{10})

60 = 10 lo g_{10} (x)

lo g_{10} (x + 1) = 2

What is the answer to log_{10}(\sqrt[5]{x^2-21x})= 2/5 ?
The answer to log_{10}(\sqrt[5]{x^2-21x})= 2/5 is x=25,x=-4
Solve for x: log_{10}(\sqrt[5]{x^2-21x})= 2/5
The solution is x=25,x=-4