What is ln(e^{2x}-1)-ln(e^{2x}+1) ?

The solution to ln(e^{2x}-1)-ln(e^{2x}+1) is ln((e^{2x}-1)/(e^{2x)+1})

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simplify ln(e^{2x}-1)-ln(e^{2x}+1)

simplify

ln (e^{2 x} - 1) - ln (e^{2 x} + 1)

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

Solution steps

ln (e^{2 x} - 1) - ln (e^{2 x} + 1)

Apply log rule:

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

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

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

ln (\frac{1}{64 + 18 - 16 ln ( 16 )})

ln (\frac{201}{200})

ln (u x^{x - n})

ln (\frac{1000}{x})

ln (\frac{1}{( 1 + s ^{w} ) ^{x + 1}})

What is ln(e^{2x}-1)-ln(e^{2x}+1) ?
The solution to ln(e^{2x}-1)-ln(e^{2x}+1) is ln((e^{2x}-1)/(e^{2x)+1})