Difference between revisions of "1954 AHSME Problems/Problem 38"
(→Solution) |
|||
Line 7: | Line 7: | ||
==Solution== | ==Solution== | ||
− | 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{( | + | 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{(B) \ } 1.47 }</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 12:43, 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 | |
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 |