If $(t-2)= e^{3(x-1)}$ then $x=?$

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If $(t-2)= e^{3(x-1)} $ then $x=?$. I guess I have to change the right side of the equation to get the x to the other side.

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Hint: Taking the natural logarithm of both sides we obtain $$\text{ln}(t-2)=\text{ln}(e^{3(x-1)})$$ $$\Leftrightarrow \text{ln}(t-2)={3(x-1)}\text{ln}(e)$$

Note that $\text{ln}(e)=1$. So we obtain

$$\Leftrightarrow \text{ln}(t-2)={3(x-1)}$$

Then you just solve for $x$ by dividing both sides by 3, adding 1 to both sides as @peterwhy suggested.