Solve:- x^y=y^x and x^2=y^3.
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x
y
=
y
x
admits as solution
x
=
y
Now applying
log
to the equations
{
x
y
=
y
x
x
2
=
y
3
-----(1)
we obtain
{
y
log
x
=
x
log
y
2
log
x
=
3
log
y
Dividing term to term we obtain the relationship
y
2
=
x
3
The solutions for (1) are obtained solving
{
x
=
y
x
2
=
y
3
and
{
x
2
=
y
3
y
2
=
x
3
so they are
⎛
⎜
⎜
⎝
x
=
0
y
=
0
x
=
1
y
=
1
x
=
27
8
y
=
9
4
⎞
⎟
⎟
⎠
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