# Exponential algebra problem

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

My proposed solution is below.

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

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

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Nice work, but all you've proven is that $x=1.5$ is a solution. To prove it's the only one, start with your third line, rewrite it as $2^{-x+1.5}=3^{2x-3}$, and take logs. – vadim123 Sep 8 '13 at 15:45

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

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You made a mistake going from the third to the fourth line. – vadim123 Sep 8 '13 at 17:46
@vadm123 I can't see any error from the third to the fourth line – miracle173 Sep 8 '13 at 20:19
@NightRa It seems that there are a lot of special tricks are necessary to solve such a system: powers of 2 on the LHS , powers of 3 on the RHS; using logarithms of base 2 and changing from logarithm of base 2 to logarithm of base 3. all of that is not necessary. – miracle173 Sep 8 '13 at 20:32
@miracle173 That is why this post exists. For different approaches to the problem. And, I fixed the question, so there is no error now. – NightRa Sep 8 '13 at 20:42

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.

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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 $x$.

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