Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

My proposed solution is below.

share|cite|improve this question

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

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

share|cite|improve this answer
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
up vote 1 down vote accepted

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

share|cite|improve this answer
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.

share|cite|improve this answer

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 $x$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.