Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

1956 AHSME Problems/Problem 20

Revision as of 15:38, 21 June 2021 by Angrybird029 (talk | contribs) (Created page with "If <math>(0.2)^x = 2</math> and <math>\log 2 = 0.3010</math>, then the value of <math>x</math> to the nearest tenth is: <math>\textbf{(A)}\ - 10.0 \qquad\textbf{(B)}\ - 0.5 \...")
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

If $(0.2)^x = 2$ and $\log 2 = 0.3010$, then the value of $x$ to the nearest tenth is:

$\textbf{(A)}\ - 10.0 \qquad\textbf{(B)}\ - 0.5 \qquad\textbf{(C)}\ - 0.4 \qquad\textbf{(D)}\ - 0.2 \qquad\textbf{(E)}\ 10.0$

Solution

$(0.2)^x = (\frac{1}{5})^x = 5^{-x}$

$\sqrt{5} \approx 2.236$

Thus, $-x \approx 0.5$, so the answer is $\boxed{\textbf{(B)}\ -0.5}$

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1956_AHSME_Problems/Problem_20&oldid=156551"