What is log_{10}(x^2-1)1+log_{10}(x-1) ?

The solution to log_{10}(x^2-1)1+log_{10}(x-1) is log_{10}((x^2-1)(x-1))

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simplify log_{10}(x^2-1)1+log_{10}(x-1)

simplify

lo g_{10} (x^{2} - 1) 1 + lo g_{10} (x - 1)

lo g_{10} ((x^{2} - 1) (x - 1))

Solution steps

lo g_{10} (x^{2} - 1) \cdot 1 + lo g_{10} (x - 1)

Apply log rule:

a lo g_{c} (b) = lo g_{c} (b^{a})

1 \cdot lo g_{10} (x^{2} - 1) = lo g_{10} ((x^{2} - 1)^{1})

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

Apply log rule:

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

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

= lo g_{10} ((x^{2} - 1)^{1} (x - 1))

Apply rule

a^{1} = a

(x^{2} - 1)^{1} = x^{2} - 1

= lo g_{10} ((x^{2} - 1) (x - 1))

4 + 4 ln (x) - ln^{2} (x)

2 (ln (2 + 1) - 2 ln (2 - 1))

lo g_{9} ((\frac{x}{y})^{4})

- lo g_{10} (5.7 (1 0^{- 8}))

- lo g_{10} (9.49 \cdot 1 0^{- 1})

What is log_{10}(x^2-1)1+log_{10}(x-1) ?
The solution to log_{10}(x^2-1)1+log_{10}(x-1) is log_{10}((x^2-1)(x-1))