Question 1034317
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Please help me solve this question

Solve the system; logx base a.log ( xyz ) base a = 48,
                  Logy base a.log ( xyz ) base a = 12,
                  Logz base a.log ( xyz ) base a = 84,
a > 0, a is not = to 1
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<pre>
{{{log(a,(x))*log(a, (xyz))}}} = {{{48}}},   (1)
{{{log(a,(y))*log(a, (xyz))}}} = {{{12}}},   (2)
{{{log(a,(z))*log(a, (xyz))}}} = {{{84}}}.   (3)

Add (1), (2) and (3) (both sides). You will get

{{{(log(a, (xyz)))^2}}} = 48 + 12 + 84 = 144.

Hence, {{{log(a, (xyz))}}} = +/- 12.

Now, consider two cases.

a) {{{log(a, (xyz))}}} = + 12.

   Then {{{log(a,(x))}}} = {{{48/12}}} = 4  --->  x = {{{a^4}}},
        {{{log(a,(y))}}} = {{{12/12}}} = 1  --->  y = {{{a}}},
        {{{log(a,(z))}}} = {{{84/12}}} = 7  --->  z = {{{a^7}}}.

b) {{{log(a, (xyz))}}} = - 12.

   Then {{{log(a,(x))}}} = {{{-48/12}}} = -4  --->  x = {{{a^(-4)}}},
        {{{log(a,(y))}}} = {{{-12/12}}} = -1  --->  y = {{{a^(-1)}}},
        {{{log(a,(z))}}} = {{{-84/12}}} = -7  --->  z = {{{a^(-7)}}}.

Thus you have two solutions (or two sets of solutions).
</pre>