Can you solve these logarithmic equations?

$\log_{4} a^{2}bc=2 \to a^2bc=16$ ............. $(1)$

$\log_{9} ab^{2}c=2\to ab^2c=81$ .................. $(2)$

$\log_{16} abc^{2}=2\to abc^2=256$ ............... $(3)$

$(1)/(2)$ $\quad \quad$ $b=\frac{81}{16}a$

$(1)/(3)$ $\quad \quad$ $c=16a$

putting values of $b$ and $c$

$a=\frac{2}{3}$

so $b=\frac{27}{8}$ and $c=\frac{32}{3}$

The given equations are

The first equation can be written as

$\log_{2}{a}+\log_{4}{b}+\log_{4}{c}=\log_{4}{a^2ba}=2 \Rightarrow {a^2}bc=2^4............\boxed{1}$

Similarly,

$\log_{3}{b}+\log_{9}{c}+\log_{9}{a}=2 =\log_{9}{{b^2}ac} \Rightarrow {b^2}ca = 3^4............\boxed{2}$

Similarly,

$\log_{4}{c}+\log_{16}{a}+\log_{16}{b}=2=\log_{16}{{c^2}ab} \Rightarrow{c^2}ab=16^2=4^4...\boxed{3}$

Multiplying $1,2,3$ ,we get,

$(abc)^{4}=2^4 \times 3^4 \times 4^4 = (24)^4 \Rightarrow \boxed{abc=24}$

Then $1$ becomes ${a^2}bc=16 \Rightarrow a(abc)=16 \Rightarrow a=\dfrac{16}{24} \Rightarrow \boxed{a=\dfrac{2}{3}}$

Similarly $2$ becomes ${b^2}ac=81 \Rightarrow b(abc)=81 \Rightarrow b=\dfrac{81}{24} \Rightarrow \boxed{b=\dfrac{27}{8}}$

Similarly $3$ becomes ${c^2}ab = 256 \Rightarrow c(abc)=256 \Rightarrow c=\dfrac{256}{24} \Rightarrow \boxed{c=\dfrac{32}{3}}$

$6a+8b-3c = 6(\dfrac{2}{3}) + 8(\dfrac{27}{8}) -3(\dfrac{32}{3})=27+4-32 = \boxed{-1}$