Logarithm System

We have that $w = x^{24} = (xyz)^{12} \Longrightarrow x^{12} = (yz)^{12}.$

Since $x,y,z$ are all real numbers greater than $1$ we can then conclude that

$x = yz \Longrightarrow z = \dfrac{x}{y},$ and thus

$\log_{w}(z) = \log_{w}(x) - \log_{w}(y) = \dfrac{1}{\log_{x}(w)} - \dfrac{1}{\log_{y}(w)} = \dfrac{1}{24} - \dfrac{1}{40} = \dfrac{1}{60}.$

Therefore $(\log_{w}(z))^{-1} = \boxed{60}.$

Direct approach:

w=x^24=y^40=x^12y^12z^12

x^24=y^40 implies x^3=y^5

x^12=(x^3)^4=(y^5)^4=y^20

W=x^12y^12z^12=y^20y^12z^12 =y^32z^12=y^40

Therefore z^12=y^40/y^32=y^8

Implies z^60=y^40=w

Therefore z=w^(1/60)

Therefore log z (base w)=1/60

implies 1/logz (base w)=60

Using the fact that: log a (b) = 1/ log b (a) we can write:

log w (xyz) = 1/ log xyz (w) = 1/12

log w (xyz) = log w (x) + log w (y) + log w (z) = 1/24 + 1/40 + log w (z)

Combining the 2 above, we get

1/12 = 1/24 + 1/40 + log w (z)

1/log w (z) = 60

Should this be a level 4 question?

$log_x w=24$ ...(A)

$log_yw=40$ ...(B)

$log_{xyz}w=12$ ...(C)

From (C)

$(xyz)^{12}=w$ $x^{12}y^{12}z^{12}=w$ $(x^{24})^{\frac{1}{2}}(y^{40})^{\frac{3}{10}}z^{12}=w$

From (A) it follows that $x^{24}=w$ and from (B) it follows that $y^{40}=w$

$w^{\frac{1}{2}}w^{\frac{3}{10}}z^{12}=w$ $w^{\frac{4}{5}}z^{12}=w$ $z^{12}=w^{\frac{1}{5}}$ $z^{60}=w$

Therefore, $log_zw=60$ and as $log_ab=\frac{1}{log_ba}$ therefore $(log_wz)^{-1}=log_zw=\boxed{60}$