The base of a logarithm should always be positive and can't be equal to 1, so the domain is 0 < x < 1 or x > 1.
[tex]\log_{\frac1x}243=5[/tex]
Write both sides as powers of 1/x :
[tex]\left(\dfrac1x\right)^{\log_{\frac1x}243}=\left(\dfrac1x\right)^5[/tex]
Recall that [tex]a^{\log_ab}=b[/tex], so that
[tex]243=\left(\dfrac1x\right)^5[/tex]
[tex]243=\dfrac1{x^5}[/tex]
[tex]x^5=\dfrac1{243}[/tex]
Take the 5th root of both sides, recalling that 3⁵ = 243, so
[tex]x=\sqrt[5]{\dfrac1{243}}=\boxed{\dfrac13}[/tex]
