Difference between revisions of "1988 AIME Problems/Problem 3"
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<math>b^{1/3}=\frac{b}{3}\ b^{-2/3}=\frac13\ b=3^{3/2}\ \b^2=3^3=\boxed{27}</math> | <math>b^{1/3}=\frac{b}{3}\ b^{-2/3}=\frac13\ b=3^{3/2}\ \b^2=3^3=\boxed{27}</math> | ||
− | + | ~thedodecagon | |
------ | ------ | ||
− | Note that we use the property <math>\ | + | Note that we use the property <math>\log_{x^n}y=\frac1n\log_xy</math> in step 1 and <math>\frac{\log_wx}{\log_wy}=\log_yx</math> in step 2 in this solution. |
== See also == | == See also == |
Latest revision as of 20:52, 18 November 2020
Contents
[hide]Problem
Find if
.
Solution 1
Raise both as exponents with base 8:
A quick explanation of the steps: On the 1st step, we use the property of logarithms that . On the 2nd step, we use the fact that
. On the 3rd step, we use the change of base formula, which states
for arbitrary
.
Solution 2: Substitution
We wish to convert this expression into one which has a uniform base. Let's scale down all the powers of 8 to 2.
Solving, we get
, which is what we want.
Just a quick note-
In this solution, we used 2 important rules of logarithm:
1) .
2)
.
Solution 3
First we have
Changing the base in the numerator yields
Using the property
yields
Now setting
, we have
Solving gets
.
~ Nafer
Solution 4
Say that and
so we have
. And we want
.
Because (as
and
from our setup), we have that
~thedodecagon
Note that we use the property in step 1 and
in step 2 in this solution.
See also
1988 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
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.