Difference between revisions of "1983 AIME Problems/Problem 1"
5849206328x (talk | contribs) m |
Coolmath2017 (talk | contribs) (→Solution 6) |
||
(48 intermediate revisions by 11 users not shown) | |||
Line 1: | Line 1: | ||
== Problem == | == Problem == | ||
− | Let <math>x</math>, <math>y</math> | + | Let <math>x</math>, <math>y</math> and <math>z</math> all exceed <math>1</math> and let <math>w</math> be a positive number such that <math>\log_x w = 24</math>, <math>\log_y w = 40</math> and <math>\log_{xyz} w = 12</math>. Find <math>\log_z w</math>. |
− | == | + | == Solution == |
=== Solution 1 === | === Solution 1 === | ||
− | The [[logarithm]]ic notation doesn't tell us much, so we'll first convert everything to the equivalent exponential | + | The [[logarithm]]ic notation doesn't tell us much, so we'll first convert everything to the equivalent exponential forms. |
<math>x^{24}=w</math>, <math>y^{40}=w</math>, and <math>(xyz)^{12}=w</math>. If we now convert everything to a power of <math>120</math>, it will be easy to isolate <math>z</math> and <math>w</math>. | <math>x^{24}=w</math>, <math>y^{40}=w</math>, and <math>(xyz)^{12}=w</math>. If we now convert everything to a power of <math>120</math>, it will be easy to isolate <math>z</math> and <math>w</math>. | ||
Line 14: | Line 14: | ||
=== 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 containing <math>z</math> is <math>(xyz)^{12}=w</math>. It now becomes clear that 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^{\frac{1}{2}}</math>. Raising both sides of <math>y^{40}=w</math> to the <math>\frac{12}{40}</math>th power gives <math>y^{12}=w^{\frac{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}w^{3/10}z^{12}=w</math>. Simplifying, we get <math>z^{60}=w</math>. | ||
+ | So our answer is <math>\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} \\ |
\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*}</ | + | \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}</math>. | w} = \frac 1 {12} - \frac 1 {24} - \frac 1{40} = \frac 1 {60}</math>. | ||
Line 23: | Line 32: | ||
Hence, <math> \log_z w = \boxed{060}</math>. | Hence, <math> \log_z w = \boxed{060}</math>. | ||
− | {{ | + | === Solution 4 === |
+ | Since <math>\log_a b = \frac{1}{\log_b a}</math>, the given conditions can be rewritten as <math>\log_w x = \frac{1}{24}</math>, <math>\log_w y = \frac{1}{40}</math>, and <math>\log_w xyz = \frac{1}{12}</math>. Since <math>\log_a \frac{b}{c} = \log_a b - \log_a c</math>, <math>\log_w z = \log_w xyz - \log_w x - \log_w y = \frac{1}{12}-\frac{1}{24}-\frac{1}{40}=\frac{1}{60}</math>. Therefore, <math>\log_z w = \boxed{060}</math>. | ||
+ | |||
+ | === Solution 5 === | ||
+ | |||
+ | If we convert all of the equations into exponential form, we receive <math>x^{24}=w</math>, <math>y^{40}=w</math>, and <math>(xyz)^{12}=w</math>. The last equation can also be written as <math>x^{12}y^{12}z^{12}=w</math>. Also note that by multiplying the first two equations, we get, <math>x^{24}y^{40}= w^{2}</math>. Taking the square root of this, we find that <math>x^{12}y^{20}=w</math>. Recall, <math>x^{12}y^{12}z^{12}=w</math>. Thus, <math>z^{12}= y^{8}</math>. Also recall, <math>y^{40}=w</math>. Therefore, <math>z^{60}</math> = <math>y^{40}</math> = <math>w</math>. So, <math>\log_z w</math> = <math>\boxed{060}</math>. | ||
+ | |||
+ | -Dhillonr25, Bobbob | ||
+ | === Solution 6 === | ||
+ | Converting all of the logarithms to exponentials gives <math>x^{24} = w, y^{40} =w,</math> and <math>x^{12}y^{12}z^{12}=w.</math> | ||
+ | Thus, we have <math>y^{40} = x^{24} \Rightarrow z^3=y^2.</math> | ||
+ | We are looking for <math>\log_z 3,</math> which by substitution, is <math>\log_{y^{\frac{2}{3}}} y^{40} = 40 \div \frac{2}{3} =\boxed{60}.</math> | ||
+ | |||
+ | ~coolmath2017 | ||
== See Also == | == See Also == |
Revision as of 17:38, 18 December 2020
Contents
Problem
Let ,
and
all exceed
and let
be a positive number such that
,
and
. Find
.
Solution
Solution 1
The logarithmic notation doesn't tell us much, so we'll first convert everything to the equivalent exponential forms.
,
, 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 containing
is
. It now becomes clear that one way to find
is to find what
and
are in terms of
.
Taking the square root of the equation results in
. Raising both sides of
to the
th power gives
.
Going back to , we can substitute the
and
with
and
, respectively. We now have
. Simplifying, we get
.
So our answer is
.
Solution 3
Applying the change of base formula,
Therefore,
.
Hence, .
Solution 4
Since , the given conditions can be rewritten as
,
, and
. Since
,
. Therefore,
.
Solution 5
If we convert all of the equations into exponential form, we receive ,
, and
. The last equation can also be written as
. Also note that by multiplying the first two equations, we get,
. Taking the square root of this, we find that
. Recall,
. Thus,
. Also recall,
. Therefore,
=
=
. So,
=
.
-Dhillonr25, Bobbob
Solution 6
Converting all of the logarithms to exponentials gives and
Thus, we have
We are looking for
which by substitution, is
~coolmath2017
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.