Difference between revisions of "1964 AHSME Problems/Problem 1"

(Created page with "We do some simple logarithms here: <math>[\log_{10}(5\log_{10}100)]^2 = [\log_{10}(5\cdot 2)]^2 = [\log_{10}(10)]^2 = [1]^2 = 1 \rightarrow \fbox{E}</math>.")
 
(See Also)
 
(One intermediate revision by the same user not shown)
Line 1: Line 1:
We do some simple logarithms here:
+
==Problem==
  
<math>[\log_{10}(5\log_{10}100)]^2 = [\log_{10}(5\cdot 2)]^2 = [\log_{10}(10)]^2 = [1]^2 = 1 \rightarrow \fbox{E}</math>.
+
What is the value of <math>[\log_{10}(5\log_{10}100)]^2</math>?
 +
 
 +
<math>\textbf{(A)}\ \log_{10}50 \qquad
 +
\textbf{(B)}\ 25\qquad
 +
\textbf{(C)}\ 10 \qquad
 +
\textbf{(D)}\ 2\qquad
 +
\textbf{(E)}\ 1  </math> 
 +
 
 +
 
 +
==Solution==
 +
 
 +
<math>[\log_{10}(5\log_{10}100)]^2</math>
 +
 
 +
Since <math>10^2 = 100</math>, we have <math>\log_{10} 100 = 2</math>, so:
 +
 
 +
<math>[\log_{10}(5\cdot 2)]^2</math>
 +
 
 +
Multiply:
 +
 
 +
<math>[\log_{10}(10)]^2</math>
 +
 
 +
Since <math>10^1 = 10</math>, we have <math>\log_{10} 10 = 1</math>, so:
 +
 
 +
<math>[1]^2</math>
 +
 
 +
<math>1</math>
 +
 
 +
<math>\fbox{E}</math>.
 +
 
 +
==See Also==
 +
{{AHSME 40p box|year=1964|before=First Problem|num-a=2}}
 +
 
 +
[[Category:Introductory Algebra Problems]]
 +
 
 +
{{MAA Notice}}

Latest revision as of 02:05, 23 July 2019

Problem

What is the value of $[\log_{10}(5\log_{10}100)]^2$?

$\textbf{(A)}\ \log_{10}50 \qquad \textbf{(B)}\ 25\qquad \textbf{(C)}\ 10 \qquad \textbf{(D)}\ 2\qquad \textbf{(E)}\ 1$


Solution

$[\log_{10}(5\log_{10}100)]^2$

Since $10^2 = 100$, we have $\log_{10} 100 = 2$, so:

$[\log_{10}(5\cdot 2)]^2$

Multiply:

$[\log_{10}(10)]^2$

Since $10^1 = 10$, we have $\log_{10} 10 = 1$, so:

$[1]^2$

$1$

$\fbox{E}$.

See Also

1964 AHSC (ProblemsAnswer KeyResources)
Preceded by
First Problem
Followed by
Problem 2
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. AMC logo.png