ln(1.5)/5=ln(1+k)

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Solution for ln(1.5)/5=ln(1+k) equation: 


D( k )

k+1 <= 0

k+1 <= 0

k+1 <= 0

k+1 <= 0 // - 1

k <= -1

k in (-1:+oo)

ln(1.5)/5 = ln(k+1) // - ln(k+1)

ln(1.5)/5-ln(k+1) = 0

ln(1.5)/5-ln(k+1) = 0 // - ln(1.5)/5

-ln(k+1) = -(ln(1.5)/5)

-ln(k+1) = -(ln(1.5)/5) // * -1

ln(k+1) = ln(e^(ln(1.5)/5))

k+1 = e^(ln(1.5)/5)

k-e^(ln(1.5)/5)+1 = 0 // - 1-e^(ln(1.5)/5)

k = -(1-e^(ln(1.5)/5))

k = e^(ln(1.5)/5)-1

k = e^(ln(1.5)/5)-1

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