Difference between revisions of "2005 AMC 10B Problems/Problem 17"
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==Solution using chain logarithm rule== | ==Solution using chain logarithm rule== | ||
| − | As in solution 2, we can write <math>a</math> as <math>\log_4 5</math>, <math>b</math> as <math>\log_56</math>, <math>c</math> as <math>\log_67</math>, and <math>d</math> as <math>\log_78</math>. <math>a*b*c*d</math> is equivalent to <math>(\log_4 5)*(\log_5 6)*(\log_6 7)*(\log_7 8)</math>. Note that by the logarithm chain rule, this is equivalent to <math>\log_4 8</math>, which evaluates to <math>\frac{3}{2}</math>, so <math>\boxed{B}</math> is the answer. | + | As in solution 2, we can write <math>a</math> as <math>\log_4 5</math>, <math>b</math> as <math>\log_56</math>, <math>c</math> as <math>\log_67</math>, and <math>d</math> as <math>\log_78</math>. <math>a*b*c*d</math> is equivalent to <math>(\log_4 5)*(\log_5 6)*(\log_6 7)*(\log_7 8)</math>. Note that by the logarithm chain rule, this is equivalent to <math>\log_4 8</math>, which evaluates to <math>\frac{3}{2}</math>, so <math>\boxed{B}</math> is the answer. |
| + | ~solver1104 | ||
== See Also == | == See Also == | ||
{{AMC10 box|year=2005|ab=B|num-b=16|num-a=18}} | {{AMC10 box|year=2005|ab=B|num-b=16|num-a=18}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 01:55, 6 January 2019
Contents
Problem
Suppose that 
, 
, 
, and 
. What is 
?
Solution
![\[8=7^d\]](http://latex.artofproblemsolving.com/1/a/b/1ab4a8eb9c7ca7ce2a9f764045b51ac12198001b.png)
![\[8=\left(6^c\right)^d\]](http://latex.artofproblemsolving.com/c/4/1/c419cf7d9189d3f7bbede2cbdbc262d85a0f044e.png)
![\[8=\left(\left(5^b\right)^c\right)^d\]](http://latex.artofproblemsolving.com/e/3/0/e3014c91f205dd2bfd050ccee2091a60101d1399.png)
![\[8=\left(\left(\left(4^a\right)^b\right)^c\right)^d\]](http://latex.artofproblemsolving.com/b/4/6/b46aa3a8c5b151da053c0828a8f9128197a02033.png)
![\[8=4^{a\cdot b\cdot c\cdot d}\]](http://latex.artofproblemsolving.com/e/0/9/e0960d1f3ecdf4f72705bccdc37da0c9e8d7407f.png)
![\[2^3=2^{2\cdot a\cdot b\cdot c\cdot d}\]](http://latex.artofproblemsolving.com/a/a/7/aa73d23a235ef6049458c51d803cfb14f0350063.png)
![\[3=2\cdot a\cdot b\cdot c\cdot d\]](http://latex.artofproblemsolving.com/f/d/3/fd313dfe53340397f629a50fb3ccf551a5f3255b.png)
![\[a\cdot b\cdot c\cdot d=\boxed{\mathrm{(B)}\ \dfrac{3}{2}}\]](http://latex.artofproblemsolving.com/1/5/7/15740e90d586c4c47d58d60a76c5b79ee990f5fa.png)
Solution using logarithms
We can write 
as 
, 
as 
, 
as 
, and 
as 
.
We know that 
can be rewritten as 
, so 
![\[\frac{\log5}{\log4}\cdot\frac{\log6}{\log5}\cdot\frac{\log7}{\log6}\cdot\frac{\log8}{\log7}\]](http://latex.artofproblemsolving.com/b/1/8/b18704949a868fd9230c866fcbf568cf075b2068.png)
![\[\frac{\log8}{\log4}\]](http://latex.artofproblemsolving.com/9/2/8/92829bf408d01bafda18fdfa199508f9ee4e4063.png)
![\[\frac{3\log2}{2\log2}\]](http://latex.artofproblemsolving.com/d/4/7/d47fa57a7c2a0056536f8e4a52e93344a6611ab1.png)
![\[\boxed{\frac{3}{2}}\]](http://latex.artofproblemsolving.com/3/1/4/3144473766897bd294135c6df5bdfecc71a85953.png)
Solution using chain logarithm rule
As in solution 2, we can write as , as 
, as 
, and as 
. 
is equivalent to 
. Note that by the logarithm chain rule, this is equivalent to 
, which evaluates to 
, so 
is the answer.
~solver1104
See Also
| 2005 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 16 |
Followed by Problem 18 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AMC 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
