Question 1207302
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The lengths of the sides of an equilateral triangle are log4(a), log10(b), log25(a+b) 
where A and B are positive numbers. What is the value of a/b?
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<pre>
We are given 

    {{{log(4,a)}}} = {{{log(10,b)}}} = {{{log(25,(a+b))}}}.


Let  k = {{{log(4,a)}}} = {{{log(10,b)}}} = {{{log(25,(a+b))}}}.


It means that 

    {{{4^k}}} = a,       (1)

    {{{10^k}}} = b,      (2)

    {{{25^k}}} = a + b.  (3)


It implies that

    {{{4^k}}} + {{{10^k}}} = {{{25^k}}}.


Divide both sides (all the terms) by  {{{25^k}}}.  You will get

    {{{(4/25)^k}}} + {{{(10/25)^k}}} = 1,

or

    {{{(2/5)^(2k)}}} + {{{(2/5)^k}}} = 1.    (4)


Let x = {{{(2/5)^k}}}.  Then equation (4) takes the form

    {{{x^2}}} + x = 1,

or

    {{{x^2}}} + x - 1 = 0.


Its roots are  {{{x[1]}}} = {{{(-1 + sqrt(5))/2}}},  {{{x[2]}}} = {{{(-1 - sqrt(5))/2}}}.


Our value of x is positive {{{(2/5)^k}}};  so, we consider only positive root  x = {{{(sqrt(5)-1)/2}}}.


Thus we have

    {{{(2/5)^k}}} = {{{(sqrt(5)-1)/2}}},

or

    {{{(4/10)^k}}} = {{{(sqrt(5)-1)/2}}}.


But from (1) and (2),  {{{a/b}}} = {{{(4/10)^k}}}.


Thus we proved that  {{{a/b}}} = {{{(sqrt(5)-1)/2}}}.


<U>ANSWER</U>.  {{{a/b}}} = {{{(sqrt(5)-1)/2}}} = 0.618033989  (approximately).
</pre>

Solved.