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.
