What is log_{e}(a+b)+log_{e}(a-b)-2log_{e}(c) ?

The solution to log_{e}(a+b)+log_{e}(a-b)-2log_{e}(c) is ln(((a+b)(a-b))/(c^2))

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simplify log_{e}(a+b)+log_{e}(a-b)-2log_{e}(c)

simplify

lo g_{e} (a + b) + lo g_{e} (a - b) - 2 lo g_{e} (c)

ln (\frac{( a + b ) ( a - b )}{c ^{2}})

Solution steps

lo g_{e} (a + b) + lo g_{e} (a - b) - 2 lo g_{e} (c)

Apply log rule:

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

lo g_{e} (a + b) + lo g_{e} (a - b) = lo g_{e} ((a + b) (a - b))

= lo g_{e} ((a + b) (a - b)) - 2 lo g_{e} (c)

Apply log rule:

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

2 lo g_{e} (c) = lo g_{e} (c^{2})

= lo g_{e} ((a + b) (a - b)) - lo g_{e} (c^{2})

Apply log rule:

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

lo g_{e} ((a + b) (a - b)) - lo g_{e} (c^{2}) = lo g_{e} (\frac{( a + b ) ( a - b )}{c ^{2}})

= lo g_{e} (\frac{( a + b ) ( a - b )}{c ^{2}})

Apply log rule:

lo g_{e} (x) = ln (x)

= ln (\frac{( a + b ) ( a - b )}{c ^{2}})

ln (\frac{( 2 t + 1 ) ^{8}}{( 3 t - 1 ) ^{4}})

ln (\frac{e ^{x} - e ^{- x}}{2})

ln (\frac{114}{132})

ln (\frac{7 x ^{5} - 8}{x ^{2}})

ln (\frac{c}{x})

What is log_{e}(a+b)+log_{e}(a-b)-2log_{e}(c) ?
The solution to log_{e}(a+b)+log_{e}(a-b)-2log_{e}(c) is ln(((a+b)(a-b))/(c^2))