Difference between revisions of "1951 AHSME Problems/Problem 34"

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Substitute <math>\log_{10}7=x</math> in <math> 10^{\log_{10}7} \Rightarrow 10^x=?</math>
 
Substitute <math>\log_{10}7=x</math> in <math> 10^{\log_{10}7} \Rightarrow 10^x=?</math>
 
It was already stated that <math>10^x=7</math>, so our answer is <math> \textbf{(A)}\ 7</math>
 
It was already stated that <math>10^x=7</math>, so our answer is <math> \textbf{(A)}\ 7</math>
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== See Also ==
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{{AHSME 50p box|year=1951|num-b=32|num-a=34}}
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[[Category:Introductory Algebra Problems]]

Revision as of 12:40, 2 February 2013

Problem

The value of $10^{\log_{10}7}$ is:

$\textbf{(A)}\ 7\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 10\qquad\textbf{(D)}\ \log_{10}7\qquad\textbf{(E)}\ \log_{7}10$

Solution

$\log_{10}7=x \Rightarrow 10^x=7$ Substitute $\log_{10}7=x$ in $10^{\log_{10}7} \Rightarrow 10^x=?$ It was already stated that $10^x=7$, so our answer is $\textbf{(A)}\ 7$

See Also

1951 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 32
Followed by
Problem 34
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All AHSME Problems and Solutions