Difference between revisions of "1988 AIME Problems/Problem 3"
(→Solution 3) |
Thedodecagon (talk | contribs) m (→Solution 4) |
||
(9 intermediate revisions by 3 users not shown) | |||
Line 11: | Line 11: | ||
(\log_8x)^3 &= \log_2x\\ | (\log_8x)^3 &= \log_2x\\ | ||
\left(\frac{\log_2x}{\log_28}\right)^3 &= \log_2x\\ | \left(\frac{\log_2x}{\log_28}\right)^3 &= \log_2x\\ | ||
− | (\log_2x)^2 &= (\log_28)^3 = \boxed{ | + | (\log_2x)^2 &= (\log_28)^3 = \boxed{027}\\ |
\end{align*} | \end{align*} | ||
</cmath> | </cmath> | ||
Line 44: | Line 44: | ||
First we have | First we have | ||
− | <math></math> | + | <cmath>\begin{align*} |
+ | \log_2(\log_8x)&=\log_8(\log_2x)\\ | ||
+ | \frac{\log_2(\log_8x)}{\log_8(\log_2x)}&=1 | ||
+ | \end{align*}</cmath> | ||
+ | Changing the base in the numerator yields | ||
+ | <cmath>\begin{align*} | ||
+ | \frac{3\log_8(\log_8x)}{\log_8(\log_2x)}&=1\\ | ||
+ | \frac{\log_8(\log_8x)}{\log_8(\log_2x)}&=\frac{1}{3}\\ | ||
+ | \end{align*}</cmath> | ||
+ | Using the property <math>\frac{\log_ab}{\log_ac}=\log_cb</math> yields | ||
+ | <cmath>\begin{align*} | ||
+ | \log_{\log_2x}(\log_8x)&=\frac{1}{3}\\ | ||
+ | (\log_2x)^\frac{1}{3}&=\log_8x\\ | ||
+ | \sqrt[3]{\log_2x}&=\frac{\log_2x}{3} | ||
+ | \end{align*}</cmath> | ||
+ | Now setting <math>y=\log_2x</math>, we have | ||
+ | <cmath>\sqrt[3]{y}=\frac{y}{3}</cmath> | ||
+ | Solving gets <math>y=\log_2x=3\sqrt{3}\Longrightarrow(\log_2x)^2=(3\sqrt{3})^2=\boxed{27}</math>. | ||
+ | |||
+ | ~ Nafer | ||
+ | |||
+ | == Solution 4 == | ||
+ | Say that <math>\log_{2^3}x=a</math> and <math>\log_2x=b</math> so we have <math>\log_2a=\log_{2^3}b</math>. And we want <math>b^2</math>. | ||
+ | |||
+ | <math>\\ \log_2a=\frac13 \log_{2}b \ \ \text{\tiny{(step 1)}}\\ \frac{\log_2a}{\log_{2}b}=\log_ba=\frac13\\ b^{1/3}=a.</math> | ||
+ | |||
+ | Because <math>3a=b</math> (as <math>2^{3a}=x</math> and <math>2^b=x</math> from our setup), we have that | ||
+ | |||
+ | <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>\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 21:52, 18 November 2020
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.