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**Hint:**We are given an equation in two variables which also includes a logarithmic function. Therefore, we will be solving it in two ways. The first method is to solve for t-variable which can be easily done by simply rearranging the given equation. In the second method, to solve for y-variable, we will deal with logarithms and use logarithmic properties.

**Complete step by step answer:**

We shall first solve for t-variable by making simple changes to the given equation, $\ln y=2t-3$ using basic algebra. Thus, we will add 3 on both sides of the given equation.

$\begin{align}

& \Rightarrow \ln y+3=2t-3+3 \\

& \Rightarrow \ln y+3=2t \\

\end{align}$

Now, dividing the entire equation by 2, we get

$\begin{align}

& \Rightarrow \dfrac{\ln y+3}{2}=\dfrac{2t}{2} \\

& \Rightarrow \dfrac{\ln y+3}{2}=t \\

\end{align}$

$\therefore t=\dfrac{\ln y+3}{2}$ ……………. Equation (1)

In order to solve for y-variable, we must have prior knowledge of logarithmic functions. We shall first exponentiate the given equation, $\ln y=2t-3$.

$\Rightarrow {{e}^{\ln y}}={{e}^{2t-3}}$

Also, we know that $\ln $ represents log function with base e. therefore, ln can be written as $\ln ={{\log }_{e}}$.

Thus, this implies that ${{e}^{{{\log }_{e}}y}}={{e}^{2t-3}}$.

Using this information and the property of logarithms, ${{a}^{{{\log }_{a}}b}}=b$, we get \[{{e}^{{{\log }_{e}}y}}=y~\].

$\Rightarrow y={{e}^{2t-3}}$

$\therefore y={{e}^{2t-3}}$ …………………. Equation (2)

From equations (1) and (2), we get $t=\dfrac{\ln y+3}{2}$ and $y={{e}^{2t-3}}$.

Therefore, the solution for t-variable is given as $t=\dfrac{\ln y+3}{2}$ and the solution for y-variable is given as $y={{e}^{2t-3}}$.

**Note:**Another method for solving for the y-variable was by directing taking the antilogarithm of base e on the right-hand side of the equation instead of exponentiating both the sides and then using the property of logarithmic functions. However, the usage of various properties makes the answer easy to understand and does not create any confusion.

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