1970 AHSME Problems/Problem 13

Revision as of 20:33, 13 July 2019 by Talkinaway (talk | contribs)

Problem

Given the binary operation $\star$ defined by $a\star b=a^b$ for all positive numbers $a$ and $b$. Then for all positive $a,b,c,n$, we have

$\text{(A) } a\star b=b\star a\quad\qquad\qquad\ \text{(B) } a\star (b\star c)=(a\star b) \star c\quad\\ \text{(C) } (a\star b^n)=(a \star n) \star b\quad \text{(D) } (a\star b)^n =a\star (bn)\quad\\ \text{(E) None of these}$

Solution

Let $a = 2, b = 3, c=n = 4$. If all of them are false, the answer must be $E$. If one does not fail, we will try to prove it.

For option $A$, we have $2^3 = 3^2$, which is clearly false.

For option $B$, we have $2^{81} = 8^{4}$, which is false.

For option $C$, we have $2^{81} = 16^3$, which is false.

For option $D$, we have $8^4 = 2^{12}$, which is true.

The LHS is $(a^b}^n$ (Error compiling LaTeX. ! Extra }, or forgotten $.). By the elementary definition of exponentiation, this is $a^b$ multiplied by itself $n$ times. Since each $a^b$ is actually $a$ multiplied $b$ times, the expression $(a^b}^n$ (Error compiling LaTeX. ! Extra }, or forgotten $.) is $a$ multiplied by itself $bn$ times.

The RHS is $a^{bn}$. This is $a$ multiplied by itself $bn$ times.

Thus, the LHS is always equal to the RHS, so $\fbox{D}$ is the only correct statement.

See also

1970 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 12
Followed by
Problem 14
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
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

Invalid username
Login to AoPS