Exponential algebra problem

Question

We need to solve for x: $$54\cdot 2^{2x}=72^x\cdot\sqrt{0.5}$$

My proposed solution is below.

Answer 1

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: $$-x+1.5=\frac{\log_3{3^{2x-3}}}{\log_3{2}}$$ $$-x+1.5=\frac{2x-3}{\log_3{2}}$$ $$-x(1+\frac{2}{\log_32})=-1.5-\frac{3}{\log_32}$$ $$x=1.5$$

Answer 2

$$54\cdot 2^{2x}=72^x\cdot\sqrt{0.5}$$ $$2\cdot3^3\cdot2^{2x}=2^{3x}3^{2x}\cdot(\frac{1}{2})^\frac{1}{2}$$ $$3^3\cdot2^{2x+1}=2^{3x-0.5}\cdot3^{2x}$$

Now let's compare the exponents: $$2x=3$$ $$x=1.5$$ And let's check: $$2x+1=3x-0.5$$ $$x=1.5$$

Answer 3

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.

Answer 4

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 $x$.