The area enclosed by the curves $$y = \log_e(x+e^2)$$, $$x = \log_e\left(\frac{2}{y}\right)$$, above the line $$x = \log_e 2$$ and $$y = 1$$ is

Given curves are

$$y=\log(x+e^2)$$

and

$$x=\log\left(\frac2y\right)$$

From the second curve,

$$e^x=\frac2y$$

$$y=2e^{-x}$$

Now find the intersection points of the curves.

Substituting

$$y=2e^{-x}$$

in

$$y=\log(x+e^2),$$

we check the given boundary lines.

At

$$x=\log2,$$

from

$$x=\log\left(\frac2y\right),$$

$$\log2=\log\left(\frac2y\right)$$

$$2=\frac2y$$

$$y=1$$

Thus, one corner point is

$$\left(\log2,1\right)$$

Now on the line

$$y=1,$$

from

$$y=\log(x+e^2),$$

$$1=\log(x+e^2)$$

$$x+e^2=e$$

$$x=e-e^2$$

So the bounded region is enclosed between:

- upper curve: $$y=\log(x+e^2)$$

- lower curve: $$y=1$$

- left boundary: $$x=\log2$$

- right curve: $$x=\log\left(\frac2y\right)$$

Now express the first curve as $$x$$ in terms of $$y$$.

From

$$y=\log(x+e^2),$$

$$x=e^y-e^2$$

Thus area using horizontal strips is

$$A=\int_{1}^{2e^{-\log2}}

\left[\log\left(\frac2y\right)-(e^y-e^2)\right]dy$$

Since

$$2e^{-\log2}=1,$$

the upper limit comes from intersection with $$x=\log2$$.

Now the top value of $$y$$ occurs where the two curves intersect.

Putting

$$x=\log\left(\frac2y\right)$$

in

$$y=\log(x+e^2),$$

we observe that

$$y=2$$

satisfies the equation because

$$x=\log1=0$$

and

$$\log(0+e^2)=2$$

Hence limits are from $$y=1$$ to $$y=2$$.

Therefore,

$$A=\int_{1}^{2}\left[\log\left(\frac2y\right)-e^y+e^2\right]dy$$

Now,

$$\log\left(\frac2y\right)=\log2-\log y$$

Hence,

$$A=\int_1^2(\log2-\log y-e^y+e^2)\,dy$$

Using

$$\int\log y\,dy=y\log y-y,$$

we get

$$A=\left[y\log2-(y\log y-y)-e^y+e^2y\right]_1^2$$

$$=\left[2\log2-2\log2+2-e^2+2e^2\right]-\left[\log2+1-e+e^2\right]$$

$$=2+e^2-\log2-1+e-e^2$$

$$=1+e-\log2$$

Hence, the required area is

$$\boxed{1+e-\log2}$$.