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

[Jameson]

What is the value of $x$ in the following equation?

$(ab)^2=(bc)^4=(ca)^x=abc$
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[Jameson]

Congratulations to the following members for their correct solutions:

1) mente oscura
2) kaliprasad
3) anemone
4) MarkFL

Solution (from MarkFL):

Spoiler

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:

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

\(\displaystyle 4-4x+x^2=x^2-9x+8\)

\(\displaystyle 5x=4\)

\(\displaystyle x=\frac{4}{5}\)