What is e^{1/4 (ln(2x+1)-ln(2x-1))} ?

The solution to e^{1/4 (ln(2x+1)-ln(2x-1))} is \sqrt[4]{(2x+1)/(2x-1)}

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simplify e^{1/4 (ln(2x+1)-ln(2x-1))}

simplify

e^{\frac{1}{4} (l n (2 x + 1) - l n (2 x - 1))}

Solution steps

e^{\frac{1}{4} (l n (2 x + 1) - l n (2 x - 1))}

Apply log rule:

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

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

= e^{\frac{1}{4} l n (\frac{2 x + 1}{2 x - 1})}

Apply exponent rule:

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

Apply log rule:

a^{l o g_{a} (b)} = b

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

\frac{ln ( 2.99 ) - ln ( 3 )}{( 2.99 ) - 3}

- lo g_{10} (4.17 \cdot 1 0^{- 4})

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

\frac{1}{2} - \frac{1}{3} (ln (2) - ln (8) - ln (1) + ln (5))

ln (x) - ln (y) + 1

What is e^{1/4 (ln(2x+1)-ln(2x-1))} ?
The solution to e^{1/4 (ln(2x+1)-ln(2x-1))} is \sqrt[4]{(2x+1)/(2x-1)}