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

The solution to log_{7}(3x+1)-2log_{7}(1+x) is log_{7}((3x+1)/((1+x)^2))

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

simplify

lo g_{7} (3 x + 1) - 2 lo g_{7} (1 + x)

lo g_{7} (\frac{3 x + 1}{( 1 + x ) ^{2}})

Solution steps

lo g_{7} (3 x + 1) - 2 lo g_{7} (1 + x)

Apply log rule:

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

2 lo g_{7} (x + 1) = lo g_{7} ((x + 1)^{2})

= lo g_{7} (3 x + 1) - lo g_{7} ((x + 1)^{2})

Apply log rule:

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

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

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

2 ln (x + 49) - 4 ln (x - 7)

ln (x - 5) + ln (x + 5)

ln (x^{2} - 81) - ln (x^{2} - 6 x - 27)

lo g_{b} (x^{2} y)

ln (\frac{3 x}{x ^{2} + 1})

What is log_{7}(3x+1)-2log_{7}(1+x) ?
The solution to log_{7}(3x+1)-2log_{7}(1+x) is log_{7}((3x+1)/((1+x)^2))