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Here's the Question :

If $xy$ = $64$ and $\log_x y + \log_y x = \frac{5}{2}$, find $x$ and $y$

I can get this to

$$log_x y + \frac{1}{\log_x y} \frac{5}{2}$$

let $\log_x y = N$

$$N + \frac{1}{N} = \frac{5}{2}$$

Multiply by 2

$$2N + \frac{2}{N} = 5$$

Multiply by N

$$2N^2 + 2 = 5N$$

$$2N^2 - 5N + 2 = 0$$

$$(2N - 1)(N - 2)$$

Giving : $$N = \frac{1}{2}$$

$$N = 2$$

Therefore :

$$\log_x y = \frac{1}{2} $$

$$\log_x y = 2$$

Giving

$$x^2 = y$$

$$x^{\frac{1}{2}} = y$$

Part of the original question :

$$xy = 64$$

As $x^2 = y$

$$x * x * x = 64$$

$$x^3 = 64$$

Therefore: $$x = 4$$ $$y = 16$$

I can't seem to solve for $y = x^{\frac{1}{2}}$ though

Solving for $x^{\frac{1}{2}} = y$

$$x^{\frac{1}{2}} * x^{\frac{1}{2}} = 64$$

$$x^{\frac{1}{2} + \frac{1}{2}} = 64$$ $$x= 64$$

$$xy= 64$$

$$64y= 64$$ Therefore

$$x = 64$$

$$y = 1$$

This is wrong though.

Answer :

$$(4,16) or (16,4)$$

I don't see how they got the second part. The first part makes sense but I'm not able to solve for $y = x^{\frac{1}{2}}$

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    $\begingroup$ Noticing the symmetry under a switch of x and y in the original question should lead to the conclusion of the second answer. $\endgroup$
    – Rammus
    May 19, 2015 at 12:40
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    $\begingroup$ Hint. The original problem is symmetric in $x$ and $y$, so if $(4,16)$ is a solution then ... $\endgroup$
    – Ethan Bolker
    May 19, 2015 at 12:41
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    $\begingroup$ You made a mistake in your algebra. You wrote $x^{\frac{1}{2}}\times x^{\frac{1}{2}} = 64$, but it's actually $x\times x^{\frac{1}{2}} = 64$ $\endgroup$
    – user222031
    May 19, 2015 at 12:42
  • $\begingroup$ @user222031 you're right, thankyou :) $\endgroup$
    – baxx
    May 19, 2015 at 12:43
  • $\begingroup$ @Rammus thanks. I thought that it made sense for them to be inverted, but I was fumbling getting there algebraically. Cheers! $\endgroup$
    – baxx
    May 19, 2015 at 12:44

2 Answers 2

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If $$y=x^{1/2}$$ and $$xy=64$$ then $$x^{3/2}=64 = 2^6$$ so $$x=2^{6\times 2/3} = 2^4 =16$$ and $$y=16^{1/2}=4.$$

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  • $\begingroup$ thanks, I made the mistake with $x^{\frac{1}{2}}$ before. I'm not too sure about the step that follows ``so'' though,$x = 2^{6x\frac{2}{3}} = 2^4 = 16$. I'm a bit uncertain about how the exponent was shifted from the $x$ to the $2$ in this case. I can see that $4^2 = 2^4$ and that $4 = 2^{\frac{4}{2}}$ (that might be a bad example...) I'm just not seeing how the exponent from the $x$ term is shifted to the $2$. Cheers ! $\endgroup$
    – baxx
    May 19, 2015 at 12:58
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    $\begingroup$ If $x^{3/2}=2^6$ then $\left(x^{3/2}\right)^{2/3}=\left(2^6\right)^{2/3}$, i.e. $x ^{3/2 \, \times \,2/3} = 2^{6 \, \times \,2/3}$ which means $x=2^4$ using$ \left(y^a\right)^b = y^{ab}$. $\endgroup$
    – Henry
    May 19, 2015 at 13:58
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$$\log_x(y)+\log_y(x)=\frac{\ln(y)}{\ln(x)}+\frac{\ln(x)}{\ln(y)}=\frac{\ln^2(x)+\ln^2(y)}{\ln(x)\ln(y)}=\frac{(\ln(x)+\ln(y))^2}{\ln(x)\ln(y)}-2=\frac{\ln^2(xy)}{\ln(x)\ln(y)}-2$$ Then if $xy=64$,

$$\log_x(y)+\log_y(x)=\frac{5}{2}\implies \frac{\ln^2(64)}{\ln(x)\ln(y)}=\frac{9}{2}\implies \frac{2\ln^2(64)}{9}=\ln(x)\ln(y)\underset{xy=64}{=}\ln(x)\ln(\frac{64}{x})=-\ln^2(x)+\ln(x)\ln(64)$$

and thus, if you set $X=\ln(x)$ you get $$X^2-\ln(64)X+\frac{2\ln^2(64)}{9}=0.$$

$$\Delta =\ln^2(64)-\frac{8}{9}\ln^2(64)=\frac{1}{9}\ln^2(64)$$

Therefore $$X=\frac{\ln(64)\pm\frac{1}{3}\ln(64)}{2}$$ and thus $X=\frac{1}{3}\ln(64)=\ln(\sqrt[3]{64})=\ln(4)$ ou $X=\frac{2}{3}\ln(64)=\ln(\sqrt[3]{64^2})=\ln(16).$

Therefore $x=4$ and $y=16$ or $x=16$ and $y=4$. Finally, we conclude that $(x,y)=(4,16)$ or $(x,y)=(16,4)$.

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  • $\begingroup$ thanks! I'm not really following this part : $\frac{\ln^2(x)+\ln^2(y)}{\ln(x)\ln(y)}=\frac{(\ln(x)+\ln(y))^2}{\ln(x)\ln(y)}-2=\frac{\ln^2(xy)}{\ln(x)\ln(y)}-2$ Where does the $-2$ come from there? $\endgroup$
    – baxx
    May 19, 2015 at 13:24
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    $\begingroup$ Because $$\ln^2(x)+\ln^2(y)=\underbrace{\ln^2(x)+2\ln(x)\ln(y)+\ln^2(y)}_{=(\ln(x)+\ln(y))^2}-2\ln(x)\ln(y)$$Is it right ? :-) $\endgroup$
    – Surb
    May 19, 2015 at 14:02
  • $\begingroup$ I'm sure it's right, bit dense for me to follow though I'm afraid. thanks though :) $\endgroup$
    – baxx
    May 21, 2015 at 2:32

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