Difference between revisions of "1983 AIME Problems/Problem 1"
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With some substitution, we get <math>w^5w^3z^{120}=w^{10}</math> and <math>\log_zw=\boxed{060}</math>. | With some substitution, we get <math>w^5w^3z^{120}=w^{10}</math> and <math>\log_zw=\boxed{060}</math>. | ||
| − | === Solution 2 === | + | ===Solution 2=== |
| + | First we'll convert everything to exponential form. | ||
| + | <math>x^{24}=w</math>, <math>y^{40}=w</math>, and <math>(xyz)^{12}=w</math>. The only expression with z is <math>(xyz)^{12}=w</math>. It now becomes clear one way to find <math>\log_z w</math> is to find what <math>x^{12}</math> and <math>y^{12}</math> are in terms of <math>w</math>. | ||
| + | |||
| + | Taking the square root of the equation <math>x^{24}=w</math> results in <math>x^{12}=w^{1/2}</math>. Taking the <math>12/40</math> root of <math>y^{40}=w</math> equates to <math>y^{12}=w^{3/10}</math>. | ||
| + | |||
| + | Going back to <math>(xyz)^{12}=w</math>, we can substitute the <math>x^{12}</math> and <math>y^{12}</math> with <math>w^{1/2}</math> and <math>w^{3/10}</math>, respectively. We now have <math>w^{1/2}<cmath>w^{3/10}</cmath>z^{12}=w</math>. Simplify we get <math>z^{60}=w</math>. | ||
| + | So our answer is \boxed{060}<math> | ||
| + | === Solution 3 === | ||
Applying the change of base formula, | Applying the change of base formula, | ||
<cmath>\begin{align*} \log_x w = 24 &\implies \frac{\log w}{\log x} = 24 \implies \frac{\log x}{\log w} = \frac 1 {24} \\ | <cmath>\begin{align*} \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_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{align*}</cmath> | \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{align*}</cmath> | ||
| − | Therefore, <math> \frac {\log z}{\log | + | Therefore, </math> \frac {\log z}{\log |
| − | w} = \frac 1 {12} - \frac 1 {24} - \frac 1{40} = \frac 1 {60}< | + | w} = \frac 1 {12} - \frac 1 {24} - \frac 1{40} = \frac 1 {60}<math>. |
| − | Hence, <math> \log_z w = \boxed{060} | + | Hence, </math> \log_z w = \boxed{060}$. |
{{alternate solutions}} | {{alternate solutions}} | ||
Revision as of 18:49, 5 August 2016
Problem
Let 
, 
, and 
all exceed 
, and let 
be a positive number such that 
, 
, and 
. Find 
.
Solutions
Solution 1
The logarithmic notation doesn't tell us much, so we'll first convert everything to the equivalent exponential expressions.
, 
, and 
. If we now convert everything to a power of 
, it will be easy to isolate and .
, 
, and 
.
With some substitution, we get 
and 
.
Solution 2
First we'll convert everything to exponential form.
, , and . The only expression with z is . It now becomes clear one way to find is to find what 
and 
are in terms of .
Taking the square root of the equation results in 
. Taking the 
root of equates to 
.
Going back to , we can substitute the and with 
and 
, respectively. We now have 
. Simplify we get 
.
So our answer is \boxed{060}$=== Solution 3 ===
Applying the change of base formula,
<cmath>\begin{align*} \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{align*}</cmath>
Therefore,$ (Error compiling LaTeX. Unknown error_msg) \frac {\log z}{\log
w} = \frac 1 {12} - \frac 1 {24} - \frac 1{40} = \frac 1 {60}$.
Hence,$ (Error compiling LaTeX. Unknown error_msg) \log_z w = \boxed{060}$.
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
See Also
| 1983 AIME (Problems • Answer Key • Resources) | ||
| Preceded by First Question |
Followed by Problem 2 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
