Difference between revisions of "1950 AHSME Problems/Problem 26"

Revision as of 08:34, 29 April 2012

Problem

If $\log_{10}{m}= b-\log_{10}{n}$ , then $m=$

$\textbf{(A)}\ \frac{b}{n}\qquad\textbf{(B)}\ bn\qquad\textbf{(C)}\ 10^{b}n\qquad\textbf{(D)}\ b-10^{n}\qquad\textbf{(E)}\ \frac{10^{b}}{n}$

Solution

We have $b=\log_{10}{10^b}$ . Substituting, we find $\log_{10}{m}= \log_{10}{10^b}-\log_{10}{n}$ . Using $\log{a}-\log{b}=\log{\dfrac{a}{b}}$ , the left side becomes $\log_{10}{\dfrac{10^b}{n}}$ . Because $\log_{10}{m}=\log_{10}{\dfrac{10^b}{n}}$ , $m=\boxed{\mathrm{(E) }\dfrac{10^b}{n}}$ .

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

@@ Line 10: / Line 10: @@
 ==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]]

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Difference between revisions of "1950 AHSME Problems/Problem 26"

Revision as of 08:34, 29 April 2012

Problem

Solution

See Also