AIME 1983 · 第 1 题

Solution 1

Converting the given equations to exponential form yields $x^{24}=w$, $y^{40}=w$, and $(xyz)^{12}=w$.

Next, converting these new equations so that the left side of each equation is a $120$th power yields $x^{120}=w^5$, $y^{120}=w^3$, and $(xyz)^{120}=x^{120}y^{120}z^{120} = w^{10}$.

Substituting for $x^{120}$ and $y^{120}$ yields $w^5w^3z^{120}=w^8z^{120}=w^{10}$, so $z^{120}=w^2$. Therefore, $w = z^{60}$ and $\log_z w=\boxed{060}$.

Solution 2

Converting the given equations to exponential form yields. $x^{24}=w$, $y^{40}=w$, and $(xyz)^{12}=x^{12}y^{12}z^{12} = w$.

Since $x^{24} = w$, $x = w^{\frac{1}{24}}$ and $x^{12} = w^{\frac{1}{2}}$. Also, since $y^{40} = w$, $y = w^{\frac{1}{40}}$ and $y^{12} = w^{\frac{3}{10}}$.

Substituting into $x^{12}y^{12}z^{12} = w$ yields $w^{\frac{1}{2}}w^{\frac{3}{10}}z^{12}=w^{\frac{4}{5}}z^{12}=w$. So $z^{12} = w^\frac{1}{5}$ and $w = z^{60}$. Therefore, $\log_{z} w =\boxed{060}$.

Solution 3

Applying the change of base formula,

$$\begin{aligned} \log_x w = 24 &\implies \frac{\log w}{\log x} = 24 \implies \frac{\log x}{\log w} = \frac 1 {24} \\ \log_y w = 40 &\implies \frac{\log w}{\log y} = 40 \implies \frac{\log y}{\log w} = \frac 1 {40} \\ \log_{xyz} w = 12 &\implies \frac{\log {w}}{\log {xyz}} = 12 \implies \frac{\log x +\log y + \log z}{\log w} = \frac 1 {12} \end{aligned}$$ Therefore, $\frac {\log z}{\log w} = \frac 1 {12} - \frac 1 {24} - \frac 1{40} = \frac 1 {60}$.

Hence, $\log_z w = \boxed{060}$.

Solution 4

Since $\log_a b = \frac{1}{\log_b a}$, the given conditions can be rewritten as $\log_w x = \frac{1}{24}$, $\log_w y = \frac{1}{40}$, and $\log_w xyz = \frac{1}{12}$. Since $\log_a \frac{b}{c} = \log_a b - \log_a c$, $\log_w z = \log_w xyz - \log_w x - \log_w y = \frac{1}{12}-\frac{1}{24}-\frac{1}{40}=\frac{1}{60}$. Therefore, $\log_z w = \boxed{060}$.

Solution 5

If we convert all of the equations into exponential form, we receive $x^{24}=w$, $y^{40}=w$, and $(xyz)^{12}=w$. The last equation can also be written as $x^{12}y^{12}z^{12}=w$. Also note that by multiplying the first two equations, we get, $x^{24}y^{40}= w^{2}$. Taking the square root of this, we find that $x^{12}y^{20}=w$. Recall, $x^{12}y^{12}z^{12}=w$. Thus, $z^{12}= y^{8}$. Also recall, $y^{40}=w$. Therefore, $z^{60}$ = $y^{40}$ = $w$. So, $\log_z w$ = $\boxed{060}$.

-Dhillonr25, Bobbob

Solution 6

Converting all of the logarithms to exponentials gives $x^{24} = w, y^{40} =w,$ and $x^{12}y^{12}z^{12}=w.$ Thus, we have $y^{40} = x^{24} \Rightarrow z^3=y^2.$ We are looking for $\log_z w,$ which by substitution, is $\log_{y^{\frac{2}{3}}} y^{40} = 40 \div \frac{2}{3} = \boxed{060}.$

~coolmath2017

Video Solution

https://youtu.be/8XjBNtFWWww