Difference between revisions of "1965 AHSME Problems/Problem 6"
(Created page with "== Problem 6== If <math>10^{\log_{10}9} = 8x + 5</math> then <math>x</math> equals: <math>\textbf{(A)}\ 0 \qquad \textbf{(B) }\ \frac {1}{2} \qquad \textbf{(C) }\ \frac {...") |
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==Solution== | ==Solution== | ||
Notice that <math>10^{\log_{10} 9} = 9</math>. Therefore, the condition we are looking for is <math>9=8x+5</math>, or <math>x=\frac{1}{2}</math>. <math>\text{So the answer is } \boxed{\textbf{(B)}}</math> | Notice that <math>10^{\log_{10} 9} = 9</math>. Therefore, the condition we are looking for is <math>9=8x+5</math>, or <math>x=\frac{1}{2}</math>. <math>\text{So the answer is } \boxed{\textbf{(B)}}</math> | ||
| + | |||
| + | ==See Also== | ||
| + | {{AHSME 40p box|year=1965|num-b=5|num-a=7}} | ||
| + | {{MAA Notice}} | ||
| + | |||
| + | [[Category:Introductory Algebra Problems]] | ||
Latest revision as of 16:04, 18 July 2024
Problem 6
If 
then 
equals:
Solution
Notice that 
. Therefore, the condition we are looking for is 
, or 
. 
See Also
| 1965 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 5 |
Followed by Problem 7 | |
| 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 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
