1955 AHSME Problems/Problem 17

Revision as of 09:45, 15 February 2021 by Coolmath34 (talk | contribs) (added redirect)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

If $\log x-5 \log 3=-2$, then $x$ equals:

$\textbf{(A)}\ 1.25\qquad\textbf{(B)}\ 0.81\qquad\textbf{(C)}\ 2.43\qquad\textbf{(D)}\ 0.8\qquad\textbf{(E)}\ \text{either 0.8 or 1.25}$

Solution

Definitions and Properties

$\log n$ is defined as the value of $x$ that satisfies the equation $n = 10^x$. Note that other bases can be applied as well, so $\log_b n$ would be defined as the answer to $n = b^x$

$y \log n = \log n^y$

$\log x - \log y = \log (x / y)$

Learn more properties here: [[1]]

Solving the Equation

$\log x-5 \log 3=-2$

$\log x- \log 3^5=-2$

$\log x- \log 243 =-2$

$\log x / 243 = -2$

$x/243 = 10^{-2}$

$x=\frac{243}{100}$ $x=\boxed{\textbf{(C) }2.43}$

See Also

1955 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 16
Followed by
Problem 18
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