2000 AIME2 #1: finding logarithm value
- Thread starter malick
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Do you mean, as you've posted, "log(45) = k", or do you mean "log<sub>4</sub>(5) = k"? (Same question for all the other log terms.)
Are you not familiar with the rules for manipulating logarithms?
Eliz.
_________
Edit: Ne'mind. Complete solution below.
Are you not familiar with the rules for manipulating logarithms?
Eliz.
_________
Edit: Ne'mind. Complete solution below.
Hello, malick!
How about helping us out by using some kind of notation?
Don't you realize that log420006 means "log of 420,006"
I will take a calculated GUESS at what you meant . . .
We let \(\displaystyle \,k\:=\:\log_4(5)\;\) [1]
Consider: \(\displaystyle \,\log_4(2000^6)\;=\;6\cdot\log_4(2000) \;= \;6\cdot\log_4(4^2\cdot5^3)\)
\(\displaystyle \;\;\;=\;6\left[\log_4(4^2)\,+\,\log_4(5^3)\right) \;=\;6\left[4\cdot\log_4(4)\,+\,3\cdot\log_4(5)\right] \;= \;6[4\cdot1\,+\,3\cdot k)\)
Hence, we have: \(\displaystyle \log_4(2000^6)\;=\;6(4\,+\,3k)\;\) [2]
From [1], we have: \(\displaystyle \,\log_5(4)\,=\,\frac{1}{k}\)
Consider: \(\displaystyle \,\log_5(2000^6)\;=\;6\cdot\log_5(2000)\;=\;6\cdot\log_5(5^3\cdot4^2)\)
\(\displaystyle \;\;\;= \;6\left[\log_5(5^3)\,+\,\log_5(4^2)\right] \;= \;6\left[3\cdot\log_5(5)\,+\,2\cdot\log_5(4)\right] \;= \;6\left[3\cdot1\,+\,2\cdot\frac{1}{k}\right]\)
Hence, we have: \(\displaystyle \,\log_5(2000^6)\;= \;6\left(3\,+\,\frac{2}{k}\right)\;\) [3]
Substitute [2] and [3] into the original expression:
\(\displaystyle \L\;\;\frac{2}{\log_4(2000^6)}\,+\,\frac{3}{\log_5(2000^6)} \;=\;\frac{2}{6(3k\,+\,2)}\;+\;\frac{3}{6\left(3\,+\,\frac{2}{k}\right)} \;=\;\frac{2}{6(3k\,+\,2)} \,+\,\frac{3}{6\left(\frac{3k\,+\,2}{k}\right)}\)
\(\displaystyle \L\;\;=\;\frac{2}{6(3k\,+\,2)} \,+\, \frac{3k}{6(3k\,+\,2)} \;=\;\frac{3x\,+\,2}{6(3x\,+\,2)}\;=\;\frac{1}{6}\;\) . . . There!
How about helping us out by using some kind of notation?
Don't you realize that log420006 means "log of 420,006"
I will take a calculated GUESS at what you meant . . .
We let \(\displaystyle \,k\:=\:\log_4(5)\;\) [1]
Consider: \(\displaystyle \,\log_4(2000^6)\;=\;6\cdot\log_4(2000) \;= \;6\cdot\log_4(4^2\cdot5^3)\)
\(\displaystyle \;\;\;=\;6\left[\log_4(4^2)\,+\,\log_4(5^3)\right) \;=\;6\left[4\cdot\log_4(4)\,+\,3\cdot\log_4(5)\right] \;= \;6[4\cdot1\,+\,3\cdot k)\)
Hence, we have: \(\displaystyle \log_4(2000^6)\;=\;6(4\,+\,3k)\;\) [2]
From [1], we have: \(\displaystyle \,\log_5(4)\,=\,\frac{1}{k}\)
Consider: \(\displaystyle \,\log_5(2000^6)\;=\;6\cdot\log_5(2000)\;=\;6\cdot\log_5(5^3\cdot4^2)\)
\(\displaystyle \;\;\;= \;6\left[\log_5(5^3)\,+\,\log_5(4^2)\right] \;= \;6\left[3\cdot\log_5(5)\,+\,2\cdot\log_5(4)\right] \;= \;6\left[3\cdot1\,+\,2\cdot\frac{1}{k}\right]\)
Hence, we have: \(\displaystyle \,\log_5(2000^6)\;= \;6\left(3\,+\,\frac{2}{k}\right)\;\) [3]
Substitute [2] and [3] into the original expression:
\(\displaystyle \L\;\;\frac{2}{\log_4(2000^6)}\,+\,\frac{3}{\log_5(2000^6)} \;=\;\frac{2}{6(3k\,+\,2)}\;+\;\frac{3}{6\left(3\,+\,\frac{2}{k}\right)} \;=\;\frac{2}{6(3k\,+\,2)} \,+\,\frac{3}{6\left(\frac{3k\,+\,2}{k}\right)}\)
\(\displaystyle \L\;\;=\;\frac{2}{6(3k\,+\,2)} \,+\, \frac{3k}{6(3k\,+\,2)} \;=\;\frac{3x\,+\,2}{6(3x\,+\,2)}\;=\;\frac{1}{6}\;\) . . . There!