Respuesta :

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]