What is (log_{10}(16-x^2))/(log_{10)(3x-4)} ?

The solution to (log_{10}(16-x^2))/(log_{10)(3x-4)} is log_{3x-4}(16-x^2)

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simplify (log_{10}(16-x^2))/(log_{10)(3x-4)}

simplify

\frac{lo g _{10} ( 16 - x ^{2} )}{lo g _{10} ( 3 x - 4 )}

lo g_{3 x - 4} (16 - x^{2})

Solution steps

\frac{lo g _{10} ( 16 - x ^{2} )}{lo g _{10} ( 3 x - 4 )}

Apply log rule:

\frac{lo g _{c} ( b )}{lo g _{c} ( a )} = lo g_{a} (b)

\frac{lo g _{10} ( 16 - x ^{2} )}{lo g _{10} ( 3 x - 4 )} = lo g_{3 x - 4} (16 - x^{2})

= lo g_{3 x - 4} (16 - x^{2})

lo g_{10} (1 0^{9})

(7 - y)^{2}

3 \times 16

3 (x + y)^{2}

1 \frac{1}{2} \div 2

What is (log_{10}(16-x^2))/(log_{10)(3x-4)} ?
The solution to (log_{10}(16-x^2))/(log_{10)(3x-4)} is log_{3x-4}(16-x^2)