Difference between revisions of "1956 AHSME Problems/Problem 8"

(Created page with "== Solution == Simple substitution yields <cmath>8 \cdot 2^{x} = 5^{0}</cmath> Reducing the equation gives <cmath>8 \cdot 2^{x} = 1</cmath> Dividing by 8 gives <cmath>2^{x}=\...")
 
(Solution)
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<cmath>8 \cdot 2^{x} = 1</cmath>
 
<cmath>8 \cdot 2^{x} = 1</cmath>
 
Dividing by 8 gives
 
Dividing by 8 gives
<cmath>2^{x}=\frac{1]{8}</cmath>
+
<cmath>2^{x}=\frac{1}{8}</cmath>
 
This simply gives that <math>x=-3</math>
 
This simply gives that <math>x=-3</math>
Therefore, the answer is <math>\fbox{(b) -30}</math>
+
Therefore, the answer is <math>\fbox{(b) -3}</math>

Revision as of 13:43, 31 August 2019

Solution

Simple substitution yields \[8 \cdot 2^{x} = 5^{0}\] Reducing the equation gives \[8 \cdot 2^{x} = 1\] Dividing by 8 gives \[2^{x}=\frac{1}{8}\] This simply gives that $x=-3$ Therefore, the answer is $\fbox{(b) -3}$