Exponential algebra problem We need to solve for x:
$$54\cdot 2^{2x}=72^x\cdot\sqrt{0.5}$$
My proposed solution is below.
 A: $$54\cdot 2^{2x}=72^x\cdot\sqrt{0.5}$$
$$2\cdot3^3\cdot2^{2x}=2^{3x}3^{2x}\cdot(\frac{1}{2})^\frac{1}{2}$$
$$3^3\cdot2^{2x+1}=2^{3x-0.5}\cdot3^{2x}$$
Now let's compare the exponents:
$$2x=3$$
$$x=1.5$$
And let's check:
$$2x+1=3x-0.5$$
$$x=1.5$$
A: Let's do this in another way:
$$3^3\cdot 2^{2x+1}=2^{3x-0.5}\cdot 3^{2x}$$
$$2^{-x+1.5}=3^{2x-3}$$
Let's take a log with base 2:
$$\log_{2}{2^{-x+1.5}}=\log_{2}{3^{2x-3}}$$
$${-x+1.5}=\log_{2}{3^{2x-3}}$$
Let's move to base 3:
$$-x+1.5=\frac{\log_3{3^{2x-3}}}{\log_3{2}}$$
$$-x+1.5=\frac{2x-3}{\log_3{2}}$$
$$-x(1+\frac{2}{\log_32})=-1.5-\frac{3}{\log_32}$$
$$x=1.5$$
A: You have
$$
2\cdot 3^3\cdot 2^{2x}=3^{2x}\cdot 2^{3x}\cdot 2^{-1/2}
$$
that becomes
$$
2^{1+2x-3x+1/2}=3^{2x-3}
$$
or
$$
2^{-x+3/2}=3^{2x-3}
$$
which becomes, taking logarithms (any base),
$$
(-x+3/2)\log 2=(2x-3)\log 3.
$$
This is a first degree equation, so it's
$$
x(2\log 3-\log 2)=3\log 3-\frac{3}{2}\log 2.
$$
With an obvious computation, the right hand side is
$$
3\log 3-\frac{3}{2}\log 2=\frac{3}{2}(2\log 3-\log2),
$$
so we can cancel out $2\log 3-\log 2=\log(9/2)\ne0$ and the solution is
$$
x=\frac{3}{2}.
$$
Even if the factors involving logarithms didn't cancel out, you'd have your solution.
A: I think this is a more systemactic way:
$$54\cdot 2^{2x}=72^x\cdot\sqrt{0.5}$$
Apply logarithm on both sides of the equation. For now the base of the logaritm does no really matter
$$\log{(54\cdot 2^{2x})}=\log{(72^x\cdot\sqrt{0.5})}$$
and simplify by applying  the laws of logarithm for  products and powers 
$$\log{(54)} + 2x \log{(2)}=x\log{(72)} + \log{(\sqrt{0.5})}$$
to get a  linear equation in x. Now solve this equation :
$$x=\frac{\log{(\sqrt{0.5})}-\log{(54)} }{2\log{(2)} - \log{(72)}}$$
Now you can try to simplify this expression for $x$.
