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Difference between revisions of "1956 AHSME Problems/Problem 20"

(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 \...")
 
(→‎Solution)
 
Line 3: Line 3:
 
<math>\textbf{(A)}\ - 10.0 \qquad\textbf{(B)}\ - 0.5 \qquad\textbf{(C)}\ - 0.4 \qquad\textbf{(D)}\ - 0.2 \qquad\textbf{(E)}\ 10.0</math>
 
<math>\textbf{(A)}\ - 10.0 \qquad\textbf{(B)}\ - 0.5 \qquad\textbf{(C)}\ - 0.4 \qquad\textbf{(D)}\ - 0.2 \qquad\textbf{(E)}\ 10.0</math>
 
==Solution==
 
==Solution==
−
<math>(0.2)^x = (\frac{1}{5})^x = 5^{-x}</math>
+
To be added.
−
 
 
−
<math>\sqrt{5} \approx 2.236</math>
 
−
 
 
−
Thus, <math>-x \approx 0.5</math>, so the answer is <math>\boxed{\textbf{(B)}\ -0.5}</math>
 

Latest revision as of 16:39, 21 June 2021

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

To be added.

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