Alphabet Song

Algebra Level 2

Let $9^{a}=8,8^{b}=7,7^{c}=6,6^{d}=5,5^{e}=4,$ and $4^{f}=3$ . The value of $\frac{abcdef}{3}$ is $\frac{x}{y}$ . Find $x+y$ .

The answer is 7.

2 solutions

John Aries Sarza
May 24, 2014

Start from $9^{abcdef}$

$9^{abcdef}=(9^{a})^{bcdef}=8^{bcdef} =(8^{b})^{cdef}=7^{cdef}=(7^{c})^{def} =6^{def}=(6^{d})^{ef}=5^{ef} =(5^{e})^{f}=4^{f}=3$ Thus, $9^{abcdef}=3$ Then , $3^{2abcdef}=3$ $2abcdef=1$ $abcdef=\frac{1}{2}$ $\frac{abcdef}{3}=\frac{1}{6}$

Therefore x=1 & y=6

$x+y=\boxed{7}$

I used logarithms and found the same answer :)

anoir trabelsi - 7 years ago

There is no need to use \ [ \ ] for all of your equations. Use that only for those which need to be on a new line by themselves, since other things appear to take up too much space.

I've edited your question, so you can use that as a reference.

Calvin Lin Staff - 7 years ago

Thanks Sir Calvin.

John Aries Sarza - 7 years ago

This is not a unique solution. I think it should require x, y to be smallest positive integers.

Stas Lisniak - 5 years, 7 months ago

Unnikrishnan V
May 24, 2014

$9^{abcdef} = 3$

$abcdef = \frac{1}{2}$

$\frac{abcdef}{3} = \frac{1}{6}$

$\frac{x}{y} = \frac{1}{6}$

$\boxed{x + y = 7}$

Alphabet Song

The answer is 7.

2 solutions

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