Solve. (6√6)^−x+3 = 1/6·216^2x−3

Is it [(6√6)^-x]+3=1/6·[216^2x]−3

or (6√6)^[−x+3] = 1/6·216^[2x−3]

This problem may look it involves logs but It can be solved using exponentials and indices laws.

Let's first write it in latex form so it is clearer:

$(6*\sqrt{6})^{-x+3}=\frac{1}{6}*216^{2x-3}$

Notice the bracket $(6*\sqrt{6})^{-x+3}$ Ignore the exponent right now, 

we know that: $(6*\sqrt{6})$=$(6*6^{0.5})$ 

Now back the exponent

$(6*6^{0.5})^{-x+3}$ using the law: $(a*b)^c=a^c*b^c$

$6^{-x+3}*6^{0.5*(-x+3)}$ Multiply the bracket by the half, we get, $6^{-x+3}*6^{-0.5x+1.5}$

Using the law $a^{b+c}=a^b*a^c$

$6^{-x}*6^3*6^{-0.5x}*6^{1.5}$ 

The right hand side now,

$\frac{1}{6}*216^{2x-3}$ 

We know that 

$216=6^3$

$\frac{1}{6}=6^{-1}$

$6^{-1}*6^{3(2x-3)}$ apply the same rules,

$6^{-1}*6^{6x-9}$ 

Now we have both sides simplified 

$6^{-x}*6^3*6^{-0.5x}*6^{1.5}$$=$$6^{-1}*6^{6x-9}$

Notice -x and -0.5x same base, we add exponents 

Notice 3 and 1.5, same base again so we add exponents

$6^{-1.5x}*6^{4.5}$$=$$6^{-1}*6^{6x-9}$

Now divide by $6^{-1}$

$6^{-1.5x}*6^{4.5-(-1)}$$=$$6^{-1.5x}*6^{5.5}=6^{6x-9}$

Notice $6^{6x-9}=6^{6x}*6^{-9}$

$6^{-1.5x}*6^{5.5}=6^{6x}*6^{-9}$

Divide by $6^{-9}$

$6^{-1.5x}*6^{14.5}=6^{6x}$

Divide by $6^{-1.5x}$

$6^{14.5}=6^{7.5x}$

The same base, therefore, exponents are equal (Law: $a^b=a^c,b=c$)

$14.5=7.5x$ Divide by 7.5

$x=\frac{14.5}{7.5}=1.93333333$ the 3 is recurring.