Question 1206509
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If 2ˣ = 4^y = 8^z and 1/2x + 1/4y +  1/8z = 22/7, show that
x = 7/16, y = 7/32 and z = 7/48
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<pre>
From the given part  {{{2^x}}} = {{{4^y}}} = {{{8^z}}},  we have

    {{{2^x}}} = {{{2^(2y)}}} = {{{2^(3z)}}}.


It gives us  x = 2y = 3z,  or  y = {{{x/2}}},  z = {{{x/3}}}.


Now substitute it into equation

    {{{1/(2x)}}} + {{{1/(4y)}}} + {{{1/(8z)}}} = {{{22/7}}}.


You will get

    {{{1/(2x)}}} + {{{1/(4*(x/2))}}} + {{{1/(8*(x/3))}}} = {{{22/7}}},

or

    {{{1/(2x)}}} + {{{2/(4x)}}} + {{{3/(8x)}}} = {{{22/7}}},

    {{{4/(8x)}}} + {{{4/(8x)}}} + {{{3/(8x)}}} = {{{22/7}}},

    {{{(4+4+3)/(8x)}}} = {{{22/7}}},

    {{{11/(8x)}}} = {{{22/7}}},

    x = {{{(11*7)/(22*8)}}} = {{{7/16}}}.


Then   y = {{{x/2}}} = {{{7/32}}},  z = {{{x/3}}} = {{{7/48}}},   QED.
</pre>

Solved.