Difference between revisions of "1950 AHSME Problems/Problem 26"
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<math> \textbf{(A)}\ \frac{b}{n}\qquad\textbf{(B)}\ bn\qquad\textbf{(C)}\ 10^{b}n\qquad\textbf{(D)}\ b-10^{n}\qquad\textbf{(E)}\ \frac{10^{b}}{n} </math> | <math> \textbf{(A)}\ \frac{b}{n}\qquad\textbf{(B)}\ bn\qquad\textbf{(C)}\ 10^{b}n\qquad\textbf{(D)}\ b-10^{n}\qquad\textbf{(E)}\ \frac{10^{b}}{n} </math> | ||
| − | ==Solution== | + | ==Solution 1== |
We have <math>b=\log_{10}{10^b}</math>. Substituting, we find <math>\log_{10}{m}= \log_{10}{10^b}-\log_{10}{n}</math>. Using <math>\log{a}-\log{b}=\log{\dfrac{a}{b}}</math>, the left side becomes <math>\log_{10}{\dfrac{10^b}{n}}</math>. Because <math>\log_{10}{m}=\log_{10}{\dfrac{10^b}{n}}</math>, <math>m=\boxed{\mathrm{(E) }\dfrac{10^b}{n}}</math>. | We have <math>b=\log_{10}{10^b}</math>. Substituting, we find <math>\log_{10}{m}= \log_{10}{10^b}-\log_{10}{n}</math>. Using <math>\log{a}-\log{b}=\log{\dfrac{a}{b}}</math>, the left side becomes <math>\log_{10}{\dfrac{10^b}{n}}</math>. Because <math>\log_{10}{m}=\log_{10}{\dfrac{10^b}{n}}</math>, <math>m=\boxed{\mathrm{(E) }\dfrac{10^b}{n}}</math>. | ||
| + | |||
| + | ==Solution 2== | ||
| + | adding <math>\log_{10} n</math> to both sides: | ||
| + | <cmath>\log_{10} m + \log_{10} n=b</cmath> | ||
| + | using the logarithm property: <math>\log_a {b} + \log_a {c}=\log_a{bc}</math> | ||
| + | <cmath>\log_{10} {mn}=b</cmath> | ||
| + | rewriting in exponential notation: | ||
| + | <cmath>10^b=mn</cmath> | ||
| + | <cmath>m=\boxed{\mathrm{(E) }\dfrac{10^b}{n}}</cmath> | ||
| + | ~Vndom | ||
| + | |||
| + | ==Solution 3== | ||
| + | More simply, we can just simulate the problem, if we have <math>m = 10</math>, that means the right side must be 1, so the only way we can achieve that with distinct <math>n</math>, is if <math>b = 3</math>, and <math>n = 100</math>. With this we can look through the different answer choices substituting in <math>b</math>, <math>n</math>, and <math>m</math>, and find that <math>\boxed{\mathrm{(E)}}</math> is the only one that satisfies the question. | ||
| + | |||
| + | ~Shadow-18 | ||
| + | |||
==See Also== | ==See Also== | ||
| − | {{AHSME box|year=1950|num-b=25|num-a=27}} | + | {{AHSME 50p box|year=1950|num-b=25|num-a=27}} |
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 19:38, 29 October 2020
Problem
If 
, then 
Solution 1
We have 
. Substituting, we find 
. Using 
, the left side becomes 
. Because 
, 
.
Solution 2
adding 
to both sides:
![\[\log_{10} m + \log_{10} n=b\]](http://latex.artofproblemsolving.com/a/c/d/acd4aae819e142a2d1f955adf4173b699ab685c3.png)
using the logarithm property: 
![\[\log_{10} {mn}=b\]](http://latex.artofproblemsolving.com/c/b/d/cbded8a2340f38f2e9c93e2c89067f6457b1429b.png)
rewriting in exponential notation:
![\[10^b=mn\]](http://latex.artofproblemsolving.com/8/b/1/8b10c86f05dc25ebc5f3cea860f97588d503106d.png)
![\[m=\boxed{\mathrm{(E) }\dfrac{10^b}{n}}\]](http://latex.artofproblemsolving.com/9/9/d/99d0c6f5d34ea6d45d7ce6820dc2985563f111a6.png)
~Vndom
Solution 3
More simply, we can just simulate the problem, if we have 
, that means the right side must be 1, so the only way we can achieve that with distinct 
, is if 
, and 
. With this we can look through the different answer choices substituting in 
, , and 
, and find that 
is the only one that satisfies the question.
~Shadow-18
See Also
| 1950 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 25 |
Followed by Problem 27 | |
| 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 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
