What real value of $x$ satisfies the equation $36^{x-1}=6$ ?

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$$36^{x-1} = 6$$ $$(6^2)^{x-1} = 6$$ $$6^{2x-2} = 6^1$$ $$2x-2 = 1$$ $$2x= 3$$ $$x= \boxed{\frac{3}{2}}$$

Since $\sqrt{36} = 2$, we want the exponent on the left side of the equation to equal $\dfrac{1}{2}$.

\ $$x-1 = \frac{1}{2 }$$ $$x = \boxed{\frac{3}{2}}$$