Difference between revisions of "1983 AIME Problems/Problem 1"
Coolmath2017 (talk | contribs) (→Solution 6) |
Lollipop315 (talk | contribs) m (→Solution 6) |
||
Line 43: | Line 43: | ||
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> | 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> | Thus, we have <math>y^{40} = x^{24} \Rightarrow z^3=y^2.</math> | ||
− | We are looking for <math>\log_z | + | We are looking for <math>\log_z w,</math> which by substitution, is <math>\log_{y^{\frac{2}{3}}} y^{40} = 40 \div \frac{2}{3} =\boxed{60}.</math> |
~coolmath2017 | ~coolmath2017 |
Revision as of 12:28, 28 December 2021
Contents
[hide]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.