Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

$x\approx26.799$
In exponential form, the given logarithmic equation, $\ln (2x+1)=4 ,$ is equivalent to \begin{array}{l}\require{cancel} \log_e (2x+1)=4 \\\\ 2x+1=e^4 .\end{array} Hence, the value of the variable that satisfies the given equation is \begin{array}{l}\require{cancel} 2x=e^4-1 \\\\ x=\dfrac{e^4-1}{2} \\\\ x\approx26.799 .\end{array}