Problem of the week #88 - December 2nd, 2013

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Jameson

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Congratulations to the following members for their correct solutions:

1) mente oscura
3) anemone
4) MarkFL

Solution (from MarkFL):
We are given:

$$\displaystyle (ab)^2=(bc)^4=(ca)^x=abc$$

I will assume that all of the variables are positive and not equal to one.

Begin with:

$$\displaystyle (ab)^2=(ac)^x$$

Taking the natural log of both sides, we obtain after applying the rules concerning logarithms:

$$\displaystyle 2\ln(a)+2\ln(b)=x\ln(a)+x\ln(c)$$

Now from:

$$\displaystyle (ab)^2=abc$$

We obtain:

$$\displaystyle c=ab$$

Hence:

$$\displaystyle 2\ln(a)+2\ln(b)=x\ln(a)+x\ln(a)+x\ln(b)$$

$$\displaystyle 2\ln(a)+2\ln(b)=2x\ln(a)+x\ln(b)$$

Solving for $$\displaystyle \ln(a)$$, we obtain:

(1) $$\displaystyle \ln(a)=\frac{2-x}{2(x-1)}\ln(b)$$

Next, we may use:

$$\displaystyle (bc)^4=(ac)^x$$

Taking the natural log of both sides, we obtain after applying the rules concerning logarithms:

$$\displaystyle 4\ln(b)+4\ln(c)=x\ln(a)+x\ln(c)$$

Using $$\displaystyle c=ab$$, there results:

$$\displaystyle 4\ln(b)+4\ln(a)+4\ln(b)=x\ln(a)+x\ln(a)+x\ln(b)$$

$$\displaystyle 4\ln(a)+8\ln(b)=2x\ln(a)+x\ln(b)$$

Solving for $$\displaystyle \ln(a)$$, we obtain:

(2) $$\displaystyle \ln(a)=\frac{x-8}{2(2-x)}\ln(b)$$

Using (1) and (2), we obtain:

$$\displaystyle \frac{2-x}{2(x-1)}\ln(b)=\frac{x-8}{2(2-x)}\ln(b)$$

Multiplying through by $$\displaystyle \frac{2}{\ln(b)}$$, we find:

$$\displaystyle \frac{2-x}{x-1}=\frac{x-8}{2-x}$$

Cross-multiplying, we get:

$$\displaystyle (2-x)^2=(x-1)(x-8)$$

$$\displaystyle 4-4x+x^2=x^2-9x+8$$

$$\displaystyle 5x=4$$

$$\displaystyle x=\frac{4}{5}$$

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