Difference between revisions of "1967 AHSME Problems/Problem 6"

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<math>3*4^{x}</math>
 
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<math>\fbox{D}</math>
 
<math>\fbox{D}</math>
                    --Amkan2022
 
  
 
== See also ==
 
== See also ==
{{AHSME box|year=1967|num-b=5|num-a=7}}   
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{{AHSME 40p box|year=1967|num-b=5|num-a=7}}   
  
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 00:35, 16 August 2023

Problem

If $f(x)=4^x$ then $f(x+1)-f(x)$ equals:

$\text{(A)}\ 4\qquad\text{(B)}\ f(x)\qquad\text{(C)}\ 2f(x)\qquad\text{(D)}\ 3f(x)\qquad\text{(E)}\ 4f(x)$

Solution

The desired expression is equal to $4^{x+1} - 4^{x}$ Using the fact that $4^{x+1}$=$4^{x}*4$, we see that the answer is $3*4^{x}$ $\fbox{D}$

See also

1967 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
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
All AHSME Problems and Solutions

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