MHB The solution is $x=8$.

[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:

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

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

Begin with:

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

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

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

Now from:

$$(ab)^2=abc$$

We obtain:

$$c=ab$$

Hence:

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

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

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

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

Next, we may use:

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

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

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

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

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

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

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

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

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

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

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

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

Cross-multiplying, we get:

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

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

$$5x=4$$

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