Question 372029
{{{log(4,(X))+log(8,(X))=1}}}
{{{y1+y2=1}}}
By definition,
{{{y1=log(4,(X))}}}
{{{X=4^(y1)}}}
{{{X=(2^2)^y1}}}
{{{X=2^(2*y1)}}}
.
.
{{{y2=log(8,(X))}}}
{{{X=8^(y2)}}}
{{{X=(2^3)^y2}}}
{{{X=2^(3*y2)}}}
.
.
They both equal {{{X}}} so set them equal to each other.
{{{2^(2*y1)=2^(3*y2)}}}
{{{2*y1=3*y2}}}
{{{y1=(3/2)y2}}}
Substituting,
{{{y1+y2=1}}}
{{{(3/2)y2+y2=1}}}
{{{(5/2)y2=1}}}
{{{y2=2/5}}}
{{{log(8,(X)=2/5)}}}
{{{highlight(X=8^(2/5))}}}