What is log_{10}(3x-1)+log_{10}(x-1) ?

The solution to log_{10}(3x-1)+log_{10}(x-1) is log_{10}((3x-1)(x-1))

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simplify log_{10}(3x-1)+log_{10}(x-1)

simplify

lo g_{10} (3 x - 1) + lo g_{10} (x - 1)

lo g_{10} ((3 x - 1) (x - 1))

Solution steps

lo g_{10} (3 x - 1) + lo g_{10} (x - 1)

Apply log rule:

lo g_{c} (a) + lo g_{c} (b) = lo g_{c} (ab)

lo g_{10} (3 x - 1) + lo g_{10} (x - 1) = lo g_{10} ((x - 1) (3 x - 1))

= lo g_{10} ((3 x - 1) (x - 1))

lo g_{6} (\frac{( x y ) ^{6}}{z})^{4}

ln (\frac{e ^{4 x}}{x ^{2} + 1})

ln (6.56 \cdot 1 0^{- 4})

ln (6) + ln (\frac{e ^{(2.29928)}}{3})

lo g_{2} (8^{7})

What is log_{10}(3x-1)+log_{10}(x-1) ?
The solution to log_{10}(3x-1)+log_{10}(x-1) is log_{10}((3x-1)(x-1))