exam problem.

kreshnik · **Joined:** Sun, 25 Dec 2011 00:28:25 UTC **Posts:** 95

Hello.

I've posted pretty much a similar question time ago, and according to that, here's what I've done.
Task is given:

$$1.\;\;\text{From:}\;\;\frac{4*9}{16}=2^a*3^b\;\;\;,\;\;\;\text{find:}\;\;a+b=?$

$$A)\;0$

$$B)\;2$

$$C)\;4$

$$D)\;6$

Attempted:

$$\frac{4*9}{16}=2^a*3^b$

$$\frac{9}{4}=2^a*3^b$

$$\frac{3^2}{2^2}=2^a*3^b$

$$\left(\frac{3}{2}\right)^2=2^a*3^b\;\;\;\;\;|\;\text{log}$

$$\text{log}\left(\frac{3}{2}\right)^2=\text{log}(2^a*3^b)$

$$2\;\text{log}\frac{3}{2}=\text{log}(2)^a+\text{log}(3)^b$

$$2\;\text{log}(3)-2\;\text{log}(2)=a\;\text{log}(2)+b\;\text{log}(3)$

$$-2\;\text{log}(2)+2\;\text{log}(3)=a\;\text{log}(2)+b\;\text{log}(3)$

$a=-2\;\;\;,\;\;\;b=2\;\;\;So:\;\;\;\;a+b=-2+2\;\;\Longrightarrow\;a+b=0$

I need to know if I've done it right, Thank you.

Shadow · **Posted:** Mon, 6 Feb 2012 19:33:16 UTC

This problem does not have a well-defined answer.

For example, I could choose $$a=2\log_2\left({3\over 2}\right)$ and b=0

and get another set of numbers which have a different sum.

However, it seems they wanted you to assume something like "a and b are both integers", and in that case you are safe.

Soroban · **Posted:** Mon, 6 Feb 2012 20:07:36 UTC

Hello, kreshnik!

You don't need logs for this probem . . .

Quote:

$$\text{1. Given: }\;\frac{4\cdot9}{16}\:=\:2^a3^b\qquad\text{Find: }\,a+b$

. . $(A)\;0 \qquad (B)\;2 \qquad (C)\;4 \qquad (D)\;6$

We have: . $$\frac{4\cdot9}{16} \:=\:\frac{2^2\cdot3^2}{2^4} \:=\:\frac{3^2}{2^2} \;=\;2^{\text{-}2}\cdot3^2$

. . Hence: . $a = \text{-}2,\;b = 2$

Got it?

Shadow · **Posted:** Mon, 6 Feb 2012 22:14:56 UTC

Soroban wrote:

Hello, kreshnik!

You don't need logs for this probem . . .

Quote:

$$\text{1. Given: }\;\frac{4\cdot9}{16}\:=\:2^a3^b\qquad\text{Find: }\,a+b$

. . $(A)\;0 \qquad (B)\;2 \qquad (C)\;4 \qquad (D)\;6$

We have: . $$\frac{4\cdot9}{16} \:=\:\frac{2^2\cdot3^2}{2^4} \:=\:\frac{3^2}{2^2} \;=\;2^{\text{-}2}\cdot3^2$

. . Hence: . $a = \text{-}2,\;b = 2$

Got it?

It's still not a unique choice.

mathematic · **Joined:** Fri, 1 Jul 2011 01:17:26 UTC **Posts:** 324

Shadow wrote:

Soroban wrote:

Hello, kreshnik!

You don't need logs for this probem . . .

Quote:

$$\text{1. Given: }\;\frac{4\cdot9}{16}\:=\:2^a3^b\qquad\text{Find: }\,a+b$

. . $(A)\;0 \qquad (B)\;2 \qquad (C)\;4 \qquad (D)\;6$

We have: . $$\frac{4\cdot9}{16} \:=\:\frac{2^2\cdot3^2}{2^4} \:=\:\frac{3^2}{2^2} \;=\;2^{\text{-}2}\cdot3^2$

. . Hence: . $a = \text{-}2,\;b = 2$

Got it?

It's still not a unique choice.

Why not?

Shadow · **Posted:** Tue, 7 Feb 2012 01:21:30 UTC

mathematic wrote:

Shadow wrote:

Soroban wrote:

Hello, kreshnik!

You don't need logs for this probem . . .

Quote:

$$\text{1. Given: }\;\frac{4\cdot9}{16}\:=\:2^a3^b\qquad\text{Find: }\,a+b$

. . $(A)\;0 \qquad (B)\;2 \qquad (C)\;4 \qquad (D)\;6$

We have: . $$\frac{4\cdot9}{16} \:=\:\frac{2^2\cdot3^2}{2^4} \:=\:\frac{3^2}{2^2} \;=\;2^{\text{-}2}\cdot3^2$

. . Hence: . $a = \text{-}2,\;b = 2$

Got it?

It's still not a unique choice.

Why not?

See my very first post in this topic.

jules · **Joined:** Mon, 6 Feb 2012 05:44:39 UTC **Posts:** 27

I think you have to answer this with some insight on the context of this question - I don't see where that was given in the OP.

is is from a basic course dealing with rational numbers where numbers are represented uniquely with the prime power factorization?

or is it from a more advanced course that would expect the use of logs to solve it?

we have 2 variables a and b and only one equation to work with - restricting the answer to be "integers" (ie prime power factorization) there is only going to be one answer but over the reals there are an infinite number of solutions

Shadow · **Posted:** Tue, 7 Feb 2012 09:10:16 UTC

jules wrote:

I think you have to answer this with some insight on the context of this question - I don't see where that was given in the OP.

is is from a basic course dealing with rational numbers where numbers are represented uniquely with the prime power factorization?

or is it from a more advanced course that would expect the use of logs to solve it?

we have 2 variables a and b and only one equation to work with - restricting the answer to be "integers" (ie prime power factorization) there is only going to be one answer but over the reals there are an infinite number of solutions

It's my policy not to make assumptions other than those from what forum the topic is posted in.

kreshnik · **Joined:** Sun, 25 Dec 2011 00:28:25 UTC **Posts:** 95

Hello.

Thank you very much guys, don't worry shadow we're not dealing with such difficult problems yet!
I think Soroban just did an easier way of doing it (comparing with mine :oops:

).

$$\frac{4\cdot9}{16}=\frac{2^2\cdot3^2}{2^4}=2^2\cdot3^2\cdot2^{-4}=2^{-2}\cdot3^2$

$$2^{-2}\cdot3^2=2^a\cdot3^b$

$$a=-2\;\;\;b=2$

$$a+b=0\;\;\;\;\text{So YES I got it!}$

By the way:

Code:

\cdot

Is cooler! :lol:

mathematic · **Joined:** Fri, 1 Jul 2011 01:17:26 UTC **Posts:** 324

Shadow wrote:

mathematic wrote:

Shadow wrote:

Soroban wrote:

Hello, kreshnik!

You don't need logs for this probem . . .

Quote:

$$\text{1. Given: }\;\frac{4\cdot9}{16}\:=\:2^a3^b\qquad\text{Find: }\,a+b$

. . $(A)\;0 \qquad (B)\;2 \qquad (C)\;4 \qquad (D)\;6$

We have: . $$\frac{4\cdot9}{16} \:=\:\frac{2^2\cdot3^2}{2^4} \:=\:\frac{3^2}{2^2} \;=\;2^{\text{-}2}\cdot3^2$

. . Hence: . $a = \text{-}2,\;b = 2$

Got it?

It's still not a unique choice.

Why not?

See my very first post in this topic.

I see your point. I was assuming a and b had to be integers.

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