How do they get rid of the “ln” in the second to last step to get the K’eq

How do they get rid of the “ln” in the second to last step to get the K'eq?

I tried to divide the 5.56 by the ln (5.56/ln) but it didn't give me the 260 like they got.

The correct answer and explanation is:

The confusion arises because “ln” refers to the natural logarithm, which is the inverse of the exponential function. To clarify the process for obtaining KeqK_{\text{eq}}, let’s break it down step by step:

1. Equation Setup: From the second-to-last step, we have: ΔG∘=−RTln⁡Keq\Delta G^\circ = -RT \ln K_{\text{eq}} where:
   • ΔG∘=−3.2 kJ/mol=−3200 J/mol\Delta G^\circ = -3.2 \, \text{kJ/mol} = -3200 \, \text{J/mol},
   • R=8.314 J/(mol\cdotpK)R = 8.314 \, \text{J/(mol·K)},
   • T=298 KT = 298 \, \text{K}.
2. Rearrange for KeqK_{\text{eq}}: ln⁡Keq=−ΔG∘RT\ln K_{\text{eq}} = -\frac{\Delta G^\circ}{RT}
3. Calculate ln⁡Keq\ln K_{\text{eq}}: Substitute the values: ln⁡Keq=−−32008.314×298\ln K_{\text{eq}} = -\frac{-3200}{8.314 \times 298} Simplify: ln⁡Keq=32002478.572≈1.29\ln K_{\text{eq}} = \frac{3200}{2478.572} \approx 1.29
4. Solve for KeqK_{\text{eq}}: To find KeqK_{\text{eq}}, take the exponential of both sides: Keq=eln⁡Keq=e1.29K_{\text{eq}} = e^{\ln K_{\text{eq}}} = e^{1.29} Use a calculator: Keq≈e1.29≈3.64K_{\text{eq}} \approx e^{1.29} \approx 3.64

If you provide the exact details from your question or the original ΔG∘\Delta G^\circ and parameters, I can confirm or refine this explanation further. The value you’re trying to get—like 260—might arise from different parameter values or units.