Difference between revisions of "2020 AIME I Problems/Problem 2"

(Solution 4 (Exponents > Logarithms))
(Video Solution)
Line 84: Line 84:
==Video Solution==
==See Also==
==See Also==

Latest revision as of 15:52, 3 October 2020


There is a unique positive real number $x$ such that the three numbers $\log_8{2x}$, $\log_4{x}$, and $\log_2{x}$, in that order, form a geometric progression with positive common ratio. The number $x$ can be written as $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m + n$.


Since these form a geometric series, $\frac{\log_2{x}}{\log_4{x}}$ is the common ratio. Rewriting this, we get $\frac{\log_x{4}}{\log_x{2}} = \log_2{4} = 2$ by base change formula. Therefore, the common ratio is 2. Now $\frac{\log_4{x}}{\log_8{2x}} = 2 \implies \log_4{x} = 2\log_8{2} + 2\log_8{x} \implies \frac{1}{2}\log_2{x} = \frac{2}{3} + \frac{2}{3}\log_2{x}$

$\implies -\frac{1}{6}\log_2{x} = \frac{2}{3} \implies \log_2{x} = -4 \implies x = \frac{1}{16}$. Therefore, $1 + 16 = \boxed{017}$.

~ JHawk0224

See here for a video solution:


Another video solution:


Video solution (*)


Solution 2

If we set $x=2^y$, we can obtain three terms of a geometric sequence through logarithm properties. The three terms are \[\frac{y+1}{3}, \frac{y}{2}, y.\] In a three-term geometric sequence, the middle term squared is equal to the product of the other two terms, so we obtain the following: \[\frac{y^2+y}{3} = \frac{y^2}{4},\] which can be solved to reveal $y = -4$. Therefore, $x = 2^{-4} = \frac{1}{16}$, so our answer is $\boxed{017}$.


Solution 3

Let $r$ be the common ratio. We have \[r = \frac{\log_4{(x)}}{\log_8{(2x)}} = \frac{\log_2{(x)}}{\log_4{(x)}}\] Hence we obtain \[(\log_4{(x)})(\log_4{(x)}) = (\log_8{(2x)})(\log_2{(x)})\] Ideally we change everything to base $64$ and we can get: \[(\log_{64}{(x^3)})(\log_{64}{(x^3)}) = (\log_{64}{(x^6)})(\log_{64}{(4x^2)})\] Now divide to get: \[\frac{\log_{64}{(x^3)}}{\log_{64}{(4x^2)}} = \frac{\log_{64}{(x^6)}}{\log_{64}{(x^3)}}\] By change-of-base we obtain: \[\log_{(4x^2)}{(x^3)} = \log_{(x^3)}{(x^6)} = 2\] Hence $(4x^2)^2 = x^3 \rightarrow 16x^4 = x^3 \rightarrow x = \frac{1}{16}$ and we have $1+16 = \boxed{017}$ as desired.


Solution 4 (Exponents > Logarithms)

Let $r$ be the common ratio, and let $a$ be the starting term ($a=\log_{8}{(2x)}$). We then have: \[\log_{8}{(2x)}=a, \log_{4}{(x)}=ar, \log_{2}{(x)}=ar^2\] Rearranging these equations gives: \[8^a=2x, 4^{ar}=x, 2^{ar^2}=x\] Deal with the last two equations first: Setting them equal gives: \[4^{ar}=2^{ar^2} \implies 2^{2ar}=2^{ar^2} \implies 2ar=ar^2 \implies r=2\] Using this value of $r$, substitute into the first and second equations (or the first and third, it doesn't really matter) to get: \[8^a=2x, 4^{2a}=x\] Changing these to a common base gives: \[2^{3a}=2x, 2^{4a}=x\] Dividing the first equation by 2 on both sides yields: \[2^{3a-1}=x\] Setting these equations equal to each other and removing the exponent again gives: \[3a-1=4a \implies a=-1\] Substituting this back into the first equation gives: \[8^{-1}=2x \implies 2x=\frac{1}{8} \implies x=\frac{1}{16}\] Therefore, $m+n=1+16=\boxed{017}$


Solution 5

We can relate the logarithms as follows:

\[\frac{\log_4{x}}{\log_8{(2x)}}=\frac{\log_2{x}}{\log_4{x}}\] \[\log_8{(2x)}\log_2{x}=\log_4{x}\log_4{x}\]

Now we can convert all logarithm bases to $2$ using the identity $\log_a{b}=\log_{a^c}{b^c}$:


We can solve for $x$ as follows:

\[\frac{1}{3}\log_2{(2x)}\log_2{x}=\frac{1}{4}\log_2{x}\log_2{x}\] \[\frac{1}{3}\log_2{(2x)}=\frac{1}{4}\log_2{x}\] \[\frac{1}{3}\log_2{2}+\frac{1}{3}\log_2{x}=\frac{1}{4}\log_2{x}\] We get $x=\frac{1}{16}$. Verifying that the common ratio is positive, we find the answer of $\boxed{017}$.


Solution 6

If the numbers are in a geometric sequence, the middle term must be the geometric mean of the surrounding terms. We can rewrite the first two logarithmic expressions as $\frac{1+\log_2{x}}{3}$ and $\frac{1}{2}\log_2{x}$, respectively. Therefore: \[\frac{1}{2}\log_2{x}=\sqrt{\left(\frac{1+\log_2{x}}{3}\right)\left(\log_2{x}\right)}\] Let $n=\log_2{x}$. We can rewrite the expression as: \[\frac{n}{2}=\sqrt{\frac{n(n+1)}{3}}\] \[\frac{n^2}{4}=\frac{n(n+1)}{3}\] \[4n(n+1)=3n^2\] \[4n^2+4n=3n^2\] \[n^2+4n=0\] \[n(n+4)=0\] \[n=0 \text{ and } -4\] Zero does not work in this case, so we consider $n=-4$: $\log_2{x}=-4 \rightarrow x=\frac{1}{16}$. Therefore, $1+16=\boxed{017}$.


Solution 7 (Official MAA)

By the Change of Base Formula the common ratio of the progression is\[\frac{\log_2 x}{\log_4 x} = \frac{\hphantom{m}\log_2x\hphantom{m}}{\frac{\log_2x}{\log_24}} = 2.\]Hence $x$ must satisfy\[2=\frac{\log_4 x}{\log_8 (2x)}= \frac{\log_2 x}{\log_2 4} \div \frac{\log_2(2x)}{\log_28} = \frac 32\cdot \frac{\log_2x}{1+\log_2x}.\]This is equivalent to $4 + 4\log_2x = 3\log_2x$. Hence $\log_2x = -4$ and $x = \frac{1}{16}$. The requested sum is $1+16 = 17$. See here for a video solution:


Video Solution


See Also

2020 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
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. AMC logo.png

Invalid username
Login to AoPS