How do you deal with logs with multiple bases? $\log_x W=24,\ \log_y W=40\ \log_{xyz} W=12$
Compute $\log_z W$.
Not sure how to unlock $z$ from the base of the third logarithm. Any help here would be much appreciated.
Should I use the change of base formula?
 A: Since $$\log_wxyz=\log_wx+\log_wy+\log_wz,$$ just solve the following equation:
$$\frac{1}{12}=\frac{1}{24}+\frac{1}{40}+\frac{1}{t},$$
where $t=\log_zw.$
I got $\log_zw=60.$
A: $$(xyz)^{12}=W\Rightarrow \log_z(W)=12\log_z(xy)+12\\x^{120}=W^5\text{ and } y^{120}=W^3\Rightarrow(xy)^{120}=W^8\Rightarrow \log_z(W)=15\log_z(xy)$$
Thus $$\log_z(W)=\frac{12}{15}\log_z(W)+12\Rightarrow \color{red}{\log_z(W)=12\cdot5=60}$$
A: You have $x^{24}=W$, $y^{40} = W$, and $(xyz)^{12}=W.$  Square this last equation and start substituting:
$$(xyz)^{24} = W^2$$
$$x^{24}(yz)^{24}=W^2$$
$$(yz)^{24} = W$$
$$(yz)^{120} = W^5$$
$$(y^{40})^3 z^{120} = W^5$$
$$W^3 z^{120} = W^5$$
$$z^{120} = W^2$$
$$z^{60} = W$$
$$\log_z W = 60$$
A: $W=x^{24}=y^{40}=(xyz)^{12}$ implies $x^2=xyz$ and $x^3=y^5$. The first of these implies $x=yz$, so the second implies $(yz)^3=y^5$, i.e., $z^3=y^2$. Hence $z^{60}=y^{40}=W$, or $\log_zW=60$.
A: Well if you must use change of base, then you have this:
$\require{cancel}$

$\log x = \frac{\log W}{24 }, \log y=\frac{\log W}{40}$ and $\log z=\frac{\log W-12\log xy}{12}$. 

$$\log _zW=\frac{\log W}{\log z}=\frac{12\log W}{\log W-12\log x-12\log y}$$
$$=\frac{12\log W}{\log W-12\frac{\log W}{24}-12\frac{\log W}{40}}$$
$$=\frac{12\cancel{\log W}}{(1-\frac12-\frac3{10})\cancel{\log W}}$$
$$\log _zW=\frac{12}{0.2}=60 $$
