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Step-by-Step Solution
Step 1: Identify the Dimensionless Nature of Quantity P
The given expression is
$P = \frac{\alpha}{\beta}\,\log_e\left(\frac{k\,t}{\beta\,x}\right)$.
Since $P$ is stated to be a dimensionless quantity, every factor in this product and the argument of the logarithm must also be dimensionless.
Step 2: Recognize the Logarithmic Term has to be Dimensionless
The argument of the natural logarithm,
$\displaystyle \frac{k\,t}{\beta\,x}$, must be dimensionless.
Therefore,
$$
\left[\frac{k\,t}{\beta\,x}\right] \;=\; 1
$$
where $[\dots]$ indicates “the dimension of”.
Step 3: Establish the Dimensions of k and t
Boltzmann’s constant $k$ has the dimension of energy per unit temperature.
Recall that energy (E) has dimensions $[M\,L^2\,T^{-2}]$, and temperature (T) is $[K]$ (Kelvin). Essentially,
$$
[k] = \frac{\,[M\,L^2\,T^{-2}]\,}{[K]}.
$$
When multiplied by $t$ (temperature), the result $k\,t$ has the dimension of energy:
$$
[k\,t] = [M\,L^2\,T^{-2}].
$$
Step 4: Deduce the Dimension Requirement from the Log Term
Since
$\displaystyle \frac{k\,t}{\beta\,x}$
is dimensionless, we have
$$
\frac{[M\,L^2\,T^{-2}]}{[\beta]\,[L]} \;=\; 1.
$$
Thus,
$$
[\beta] \times [L] \;=\; [M\,L^2\,T^{-2}],
$$
implying
$$
[\beta] = [M\,L\,T^{-2}].
$$
Step 5: Relate the Dimensions of α to β
In the original dimensionless quantity
$P = \frac{\alpha}{\beta}\,\log_e(\dots)$,
the factor $\displaystyle \frac{\alpha}{\beta}$ must also be dimensionless.
Therefore:
$$
\frac{[\alpha]}{[\beta]} = 1
\quad \Longrightarrow \quad
[\alpha] = [\beta].
$$
Since we found
$[\beta] = [M\,L\,T^{-2}]$,
it follows that
$$
[\alpha] = [M\,L\,T^{-2}].
$$
Step 6: Final Answer
Thus, the dimension of $\alpha$ is
$[M\,L\,T^{-2}]$.
