ln(T/298)=-2/5ln(5/20)

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Solution for ln(T/298)=-2/5ln(5/20) equation: 


D( T )

T/298 <= 0

T/298 <= 0

T/298 <= 0

1/298*T+0 <= 0 // - 0

1/298*T <= 0 // : 1/298

T <= 0/1/298

T <= 0

T in (0:+oo)

ln(T/298) = (-2/5)*ln(5/20) // - (-2/5)*ln(5/20)

ln(T/298)-((-2/5)*ln(5/20)) = 0

ln(T/298)+(2/5)*ln(5/20) = 0

ln(T/298)+2/5*ln(1/4) = 0 // - 2/5*ln(1/4)

ln(T/298) = -(2/5*ln(1/4))

ln(T/298) = -2/5*ln(1/4)

ln(T/298) = ln(e^(-2/5*ln(1/4)))

T/298 = e^(-2/5*ln(1/4))

T/298-e^(-2/5*ln(1/4)) = 0

1/298*T-e^(-2/5*ln(1/4)) = 0 // + e^(-2/5*ln(1/4))

1/298*T = e^(-2/5*ln(1/4)) // : 1/298

T = (e^(-2/5*ln(1/4)))/1/298

T = 298*e^(-2/5*ln(1/4))

T = 298*e^(-2/5*ln(1/4))

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