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

$\dfrac{1}{2}$

$1$

$\dfrac{3}{2}$

$2$

$3$

Approach 1

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

Approach 1

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

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