Difference between revisions of "1958 AHSME Problems/Problem 25"

(Solution)
(Solution)
 
Line 10: Line 10:
  
 
== Solution ==
 
== Solution ==
125. Rearrange the equation into <cmath> \log_{5}{k}\cdot\log_{k}{x} = 3</cmath>
+
125. Rearrange the equation into <cmath> \log_{5}{k}\cdot\log_{k}{x} = 3</cmath> which is just <cmath> \log_{5}{x} = 3</cmath> and x is 125.
  
 
== See Also ==
 
== See Also ==

Latest revision as of 15:16, 19 May 2024

Problem

If $\log_{k}{x}\cdot \log_{5}{k} = 3$, then $x$ equals:

$\textbf{(A)}\ k^6\qquad  \textbf{(B)}\ 5k^3\qquad  \textbf{(C)}\ k^3\qquad  \textbf{(D)}\ 243\qquad  \textbf{(E)}\ 125$

Solution

125. Rearrange the equation into \[\log_{5}{k}\cdot\log_{k}{x} = 3\] which is just \[\log_{5}{x} = 3\] and x is 125.

See Also

1958 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 24
Followed by
Problem 26
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png