Difference between revisions of "1954 AHSME Problems/Problem 38"
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==Solution== | ==Solution== | ||
− | Taking the | + | Taking the logarithm in base <math>3</math> of both sides, we get <math>x+3 = \log_3 135</math>. Using the property <math>\log ab = \log a + \log b</math>, we get <math>x+3 = \log_3 5 + \log_3 3^3</math>, or <math>x = \log_3 5</math>. Converting into base <math>10</math> gives <math>x = \frac{\log 5}{\log 3} = \frac{1 - \log 2}{\log 3}</math>. Now, plugging in the values yeilds <math>\boxed{\textbf{(C) \ } 1.67 }</math>. |
==See Also== | ==See Also== | ||
{{AHSME 50p box|year=1954|num-b=37|num-a=39}} | {{AHSME 50p box|year=1954|num-b=37|num-a=39}} |
Revision as of 12:41, 6 August 2012
Problem
If and , the value of when is approximately
Solution
Taking the logarithm in base of both sides, we get . Using the property , we get , or . Converting into base gives . Now, plugging in the values yeilds .
See Also
1954 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 37 |
Followed by Problem 39 | |
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