$(ab)^2=(bc)^4=(ca)^x=abc$ Then what is the value of $x$?

Question

Given that $(ab)^2=(bc)^4=(ca)^x=abc$ Then what is the value of $x$?

$2(\log a+\log b)=4(\log b+\log c)=x(\log c+\log a)=\log a+\log b+\log c$

Then I am lost, any other easier way to solve?

Answer 1

Hint: Split these two equalities into

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

Then use $\log$ on both equations and see what happens ;-)

Answer 2

Taking logarithm gives: $$2(\log a+\log b)=4(\log b+\log c)=x(\log c+\log a)=\log a+\log b+\log c$$

then taking $2(\log a+\log b)=\log a+\log b+\log c$

and $4(\log b+\log c)=\log a+\log b+\log c$

we would get $\log a+\log b-\log c=0$ & $3\log b+3\log c-\log a=0$ and by solving these two equations we get $\log b=-\log c$

similarly $\log a=-\log b$ then the solution becomes obvious......as $x=\frac12$

Answer 3

$2(\log a+\log b) = 4(\log b+\log c) = x(\log c+\log a ) = \log a + \log b+\log c$

Separating the equalities into two,

$$ 2(\log a+\log b)= \log a+\log b+\log c \Rightarrow \log a + \log b - \log c=0 \quad(i) $$

$$ 4(\log b+\log c)=\log a+\log b+\log c\Rightarrow 3(\log b+\log c)-\log a=0 \quad(ii) $$

From equation (i) and (ii), upon solving, we get,

$$ \log a=(3/2)\log c,~\log b=-(\log c)/2 \quad(iii) $$

$$ 4(\log b+\log c)=x(\log c+\log a) \quad(iv) $$

By putting (iii) into (iv),

$$ -2\log c+4\log c=x\log c+(3x/2)\log c, \\ 2\log c=(5x/2)\log c. $$

By dividing $\log c$ on both sides, $$ 2=(5x/2)\Rightarrow x=4/5. $$

Thanks.

Answer 4

Let’s try to solve this without logarithms, as simply as possible. This allows to extend the solution to negative and complex values of $a$,$b$ and $c$, and includes a more interesting set of solution in the specific case of $|b|=1$.

If $abc=0$, then at least two elements of $\{a,b,c\}$ are $0$, and $x$ can take any value, except $0$ if $0^0=1$.

From now on $abc\neq 0$ : $$(ab)^2=abc \Rightarrow c=ab$$ replacing $c$ by its value transforms the equalities into : $$(ab)^2=\left(ab^2\right)^4=\left(a^2b\right)^x$$ The first equality then gives $$a^2b^6=1$$ Replacing $a^2$ by $\frac1{b^6}$, we have then $$\frac1{b^4}=\left(\frac1{b^5}\right)^x \Leftrightarrow b^4=b^{5x}.$$

If $b=1$, this is true for all $x$ and $a=c=\pm1$

Otherwise, if $|b|≠1$, one has $\boxed{x=\frac45}$ and $a=\pm\frac1{b^3}$, $c=\pm\frac1{b^2}$. This is the solution you were looking for.

But, for $|b|=1$ and $b≠1$, the solution above is still true, but it is not the only one. Let’s define $β≠0$ by $b=e^{iβ}$. The conditions on $x$ then becomes $$5xβ=4β+2kπ, k∈\mathbb Z$$ giving the set of solutions $$x=\frac45+\frac{2kπ}{5β}, k∈\mathbb Z.$$

This includes, for example the non trivial solution where $a=b=i$, $c=-1$ and $x=4$.