-1
$\begingroup$

Both in a unrescrited case and with the following restriction: $a+b+c=1$

$\endgroup$
0
$\begingroup$

For the unrestricted case, let $a=1.$ Now you need $3^b7^c=5$, so for a positive integer $n$, let

$$3^b = 5^{n+1}, 7^c = 5^{-n}$$.

Solve these two equations for $b$ and $c$.

That should give you a leg up on the restricted case.

$\endgroup$
0
$\begingroup$

Hint $$c=1-a-b$$

Your equation then becomes $$\frac{10}{7}=(\frac{2}{7})^a (\frac{3}{7})^b$$

Show now that if you fix $a$ you can find some $b$ which satisfies this equation.

$\endgroup$

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.