2
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ $a,b,c$ are positive reals and $abc\ne 1$ $\endgroup$ – miosaki Nov 28 '13 at 13:27
2
$\begingroup$

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 $x=\frac12$

$\endgroup$
  • $\begingroup$ There is a mistake after "by solving these equations...": we get $2\log b=-log c$ and $\log a=-3\log b$. This also changes the value of x $\endgroup$ – Frédéric Grosshans Oct 23 '14 at 16:30
5
$\begingroup$

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 ;-)

$\endgroup$
0
$\begingroup$

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 $x=4$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.