Question 373206
{{{log(4,(x))+log(8,(x))=1}}}
{{{y1+y2=1}}}
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{{{y1=log(4,(x))}}}
{{{4^y1=x}}}
{{{(2^2)^y1=x}}}
{{{2^(2*y1)=x}}}
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{{{y2=log(8,(x))}}}
{{{8^y2=x}}}
{{{(2^3)^y2=x}}}
{{{2^(3*y2)=x}}}
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{{{2^(2*y1)=2^(3*y2)}}}
{{{2y1=3*y2)}}}
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Substitute,
{{{y1+y2=1}}}
{{{2y1+2y2=2}}}
{{{3y2+2y2=2}}}
{{{5y2=2}}}
{{{y2=2/5}}}
{{{y1=3/5}}}
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{{{x=4^(y1)=8^(y2)}}}
{{{highlight(x=root(5,4^(3))=root(5,8^(2)))}}} or approximately,
{{{x=2.2974}}}