= Solution
For the <Lagrangian sufficiency theorem>, write a maximization problem as $\max_{x\in D}f(x)$ subject to $g_i(x)\leq c_i$. Here $D$ may already incorporate simple constraints such as nonnegativity. Define the <Lagrangian in optimization>
$$
L(x,\lambda)=f(x)-\sum_i\lambda_i(g_i(x)-c_i),\qquad\lambda_i\geq0.
$$
The theorem states that a feasible point $x_*$ is a global maximizer if there are such multipliers for which $x_*$ globally maximizes $L(\cdot,\lambda)$ on $D$ and <complementary slackness> holds: $\lambda_i(g_i(x_*)-c_i)=0$ for every $i$. To prove it, for any feasible $x$ observe that
$$
f(x)\leq L(x,\lambda)\leq L(x_*,\lambda)=f(x_*).
$$
The first inequality uses feasibility and nonnegative multipliers, the second uses global maximization of the <Lagrangian in optimization>, and the last uses <complementary slackness>. Thus \b[$x_*$ is globally optimal]. No convexity hypothesis is needed for this implication; <concavity> is useful when establishing the required global maximization of $L$ from its <gradient>.
For the particular problem, put $d=e^{c_2}-1\geq0$ and $R=c_1-2d\geq0$. The logarithmic constraint is equivalent to $x_1\geq d$. The objective increases strictly with either variable, so at a maximum the resource constraint is tight: otherwise $x_1$ could be increased. Hence
$$
x_1=\frac{c_1-3x_2}{2},\qquad0\leq x_2\leq\frac R3.
$$
On this interval the objective becomes $F(x_2)=(c_1-3x_2)/2+3\log(1+x_2)$, with
$$
F'(x_2)=-\frac32+\frac3{1+x_2},\qquad F''(x_2)=-\frac3{(1+x_2)^2}<0.
$$
It increases up to $x_2=1$ and decreases thereafter, so the unique feasible maximizer is
$$
\boxed{x_2^*=\min\left\{1,\frac{c_1-2(e^{c_2}-1)}3\right\},\qquad x_1^*=\frac{c_1-3x_2^*}{2}.}
$$
The maximum value is
$$
\boxed{\begin{cases}
\displaystyle d+3\log(1+R/3),&0\leq R\leq3,\\[3pt]
\displaystyle\frac{c_1-3}{2}+3\log2,&R\geq3.
\end{cases}}
$$
The two expressions agree at $R=3$. This includes the degenerate feasible set $R=0$, where $x_1^*=d$ and $x_2^*=0$.
To certify the result using the <Lagrangian sufficiency theorem>, take $D=[0,\infty)^2$ and
$$
L=x_1+3\log(1+x_2)-\lambda(2x_1+3x_2-c_1)+\mu(\log(1+x_1)-c_2),
$$
where $\lambda,\mu\geq0$. Set
$$
\lambda=\frac1{1+x_2^*},\qquad\mu=(2\lambda-1)(1+x_1^*).
$$
Since $x_2^*\leq1$, $\lambda\geq1/2$ and $\mu\geq0$. These choices make both components of $\nabla_xL$ vanish at $x_*$: $1-2\lambda+\mu/(1+x_1^*)=0$ and $3/(1+x_2^*)-3\lambda=0$. The <Lagrangian in optimization> is <concave> on $D$, so the zero <gradient> proves global maximization there. The resource constraint is tight. When $R<3$ the logarithmic constraint is tight; when $R\geq3$, $\mu=0$. Thus <complementary slackness> holds in every case, including both endpoints, and the theorem establishes the claimed optimum.
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