Question 1141585
<pre>{{{system(log((b))/log((a))=3/2,log((d))/log((c))=5/4,a-c=9)}}}, where0 a, b, c, d are positive integers.

{{{system(3log((a))=2log((b)),5log((c))=4log((d)),a-c=9)}}}

{{{system(log((a^3))=2log((b^2)),log((c^5))=log((d^3)),a-c=9)}}}

{{{system(a^3=b^2,c^5=d^3,a-c=9)}}}

Let {{{a=p^2}}},{{{b=p^3}}},{{{c=q^3}}},{{{d=q^5}}} and perhaps
p and q can be positive integers.  We will see.

{{{system((p^2)^3=(p^3)^2,(q^4)^5=(q^4)^3,p^2-q^4=9)}}}

{{{system(p^6=p^6,q^20=q^20,p^2-q^4=9)}}}

{{{p^2-q^4=9}}}

Factor the left side:

{{{(p-q^2)(p+q^2)=9}}}

If those expressions in parentheses can be positive integers,
they can only be the unequal pair of factors of 9, which are
1 and 9. The p-q² is smaller, so it would be 1 and p+q² is
larger so it would be 9:

{{{p-q^2=1}}}, {{{p+q^2=9}}}

{{{system(p-q^2=1,p+q^2=9)}}}

Add the equations:

{{{2p=10}}}
{{{p=5}}}

Subtract the equations:

{{{-2q^2=-8}}}
{{{q^2=4}}}
{{{q=2}}}

So we now see that p and q can be positive integers.

{{{a=p^2}}},{{{b=p^3}}},{{{c=q^4}}},{{{d=q^5}}}

{{{a=5^2}}},{{{b=5^3}}},{{{c=2^4}}},{{{d=2^5}}}

{{{a=25}}},{{{b=125}}},{{{c=16}}},{{{d=32}}}

{{{b-d=125-32}}}

{{{b-d=93}}}

Edwin</pre>