Expressing \( \ln I-\ln (2 L)=\ln (K R+t)-\ln V+K V \) without logarithms gives:

To express the equation \( \ln I - \ln (2 L) = \ln (K R + t) - \ln V + K V \) without logarithms, we can use the property of logarithms that states \( \ln a - \ln b = \ln \left( \frac{a}{b} \right) \) and \( \ln a + \ln b = \ln(ab) \). Starting with the left side: \[ \ln \left( \frac{I}{2 L} \right) \] And for the right side, we can combine the logarithms: \[ \ln \left( \frac{K R + t}{V} \right) + K V \] Now, we can exponentiate both sides to eliminate the logarithm: \[ \frac{I}{2 L} = \frac{K R + t}{V} e^{K V} \] Finally, multiplying through by \( 2L \) gives: \[ I = 2L \left( \frac{K R + t}{V} e^{K V} \right) \] So the expression without logarithms is: \[ I = \frac{2L (K R + t)}{V} e^{K V} \]