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

(Solution)
(Solution)
Line 9: Line 9:
 
<math>3*4^{x}</math>
 
<math>3*4^{x}</math>
 
<math>\fbox{D}</math>
 
<math>\fbox{D}</math>
 +
                    --Amkan2022
  
 
== See also ==
 
== See also ==

Revision as of 22:02, 27 March 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 $3*4^{x}$ $\fbox{D}$

                   --Amkan2022

See also

1967 AHSME (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
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