Art of Problem Solving

1967 AHSME Problems/Problem 6: Difference between revisions

Timneh (talk | contribs)
Created page with "== Problem == If <math>f(x)=4^x</math> then <math>f(x+1)-f(x)</math> equals: <math> \text{(A)}\ 4\qquad\text{(B)}\ f(x)\qquad\text{(C)}\ 2f(x)\qquad\text{(D)}\ 3f(x)\qquad\text{..."
 
Amkan2022 (talk | contribs)
Line 5: Line 5:


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



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 $3*4^{x}$ $\fbox{D}$

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

These problems are copyrighted © by the Mathematical Association of America.