If the area above x-axis, bounded by the curves $y = 3^{\beta x}$, $x = 0$ and $x = 3$ is $\frac{26}{\log_e 3}$, then the value of $\beta$ is:

The curve is $y = 3^{\beta x}$, bounded by $x = 0$ and $x = 3$, with area $\frac{26}{\log_e 3}$.

The area under the curve is given by:

$\text{Area} = \int_0^3 3^{\beta x} \, dx$

---

Using the integration formula $\int a^{mx} \, dx = \frac{a^{mx}}{m \ln a}$:

$\int 3^{\beta x} \, dx = \frac{3^{\beta x}}{\beta \ln 3}$

where $\log_e 3 = \ln 3$

---

Evaluating the definite integral:

$\int_0^3 3^{\beta x} \, dx = \left[\frac{3^{\beta x}}{\beta \ln 3}\right]_0^3$

$= \frac{3^{3\beta}}{\beta \ln 3} - \frac{3^{0}}{\beta \ln 3}$

$= \frac{3^{3\beta} - 1}{\beta \ln 3}$

---

Setting this equal to the given area:

$\frac{3^{3\beta} - 1}{\beta \ln 3} = \frac{26}{\ln 3}$

$\frac{3^{3\beta} - 1}{\beta} = 26$

$3^{3\beta} - 1 = 26\beta$

$3^{3\beta} = 26\beta + 1$

---

For $\beta = 1$:

Left side: $3^{3(1)} = 3^3 = 27$

Right side: $26(1) + 1 = 27$

Therefore, $\beta = 1$