# Homework Help: Proof using logarithm

1. Nov 12, 2009

### songoku

1. The problem statement, all variables and given/known data
Positive integers a and b, where a < b, satisfy the equation :
ab = ba

By first taking logarithms, show that there is only one value of a and b that satisfies the equation and find the value !

2. Relevant equations
logarithm

3. The attempt at a solution
I know the solution just by guessing, a = 2 and b = 4 but I don't know how to do it...

$$a^b=b^a$$

$$b*log~a=a*log~b$$

Then I stuck....

Thanks

2. Nov 12, 2009

### rock.freak667

try taking logs to base a or b instead of base 10

3. Nov 12, 2009

### songoku

Hi rock.freak667
I think it will be the same.

$$a^b=b^a$$

$$b*\log_{a}~a=a*\log_{a}b$$

$$b=a*\log_{a}b$$

Then stuck again...:(

Thanks

4. Nov 17, 2009

### turin

Think of only one variable at a time (e.g. b in the last equation of your previous post).

5. Nov 18, 2009

### icystrike

Since a and b are positive integer,

we can conclude a,b>0 without generalisation.

First by taking log base a,

$$b=a log_{a}b$$

$$b log_{b}a=a$$

substituting to initial equation,

$$a^{2}-alog_{a}b=0$$

$$a=0$$ or $$a=log_{a}b$$

Therefore, $$a=log_{a}b$$

is the only solution for a since logarithm curve is constantly decreasing.

$$b=a^{a}$$

since exponent curve is ... leaving this part to you (=

Last edited: Nov 18, 2009
6. Nov 18, 2009

### songoku

Hi turin and icystrike
sorry I don't understand what you mean. From my last equation :

$$b=a*\log_{a}b$$

Then, think only b as variable. how to continue? what about a?

I don't understand this part. To which initial equation do you substitute?

even though I can reach this part, I still don't know how to continue. I think logarithm curve is constantly increasing, not decreasing. From y = log x, the value of y will increase if x increases.

exponent curve is also constantly increasing, but from b=aa, how to deduce that a = 2?

Thanks a lot

Last edited: Nov 18, 2009
7. Nov 19, 2009

### Staff: Mentor

I'm with Songoku on this; I'm not following what you are doing. I understand how you got both equations above, but your description is that you are taking the log base a of both sides of the original equation. In your second equation you're taking the log base b of both sides of the original equation.
Now this doesn't make any sense to me. Where did it come from? This is equivalent to a2 - b = 0 from the first of your two equations above, where you have b = a logab.
Assuming that the base is larger than 1, any log curve is constantly increasing. IOW, if x1 < x2, then logax1 < logax2.

8. Nov 19, 2009

### turin

Whoops, I thought I knew how to do the problem, until you pointed out this assumption.

9. Nov 19, 2009

### Staff: Mentor

So unless icystrike can explain what he's done, we're back at square 1 on this problem.

10. Nov 20, 2009

### icystrike

Sorry guys i've made a mistake. we are back to square one.
Now i'm able to solve it.
But it contradicts the statement that a>b.

$$((lg a)/(lg b))=((lg b)/(lg a))$$
$$(lg a)^{2}-(lg b)^{2}=0$$
$$(lg a-lg b)(lg a+lg b)=0$$
Therefore a=b which contradicts a>b
Thus the only possibility is $$a=(1/b)$$
since a is a interger,
b must divide 1,
whereby b=1
suggest a=1 .
and resulting a=b.
Which again leads to a contradiction

Last edited: Nov 20, 2009
11. Nov 20, 2009

### Staff: Mentor

How do you get the equation above? We know that ab = ba, so b loga = a logb
==> b/a = (log b)/(log a)

How do you get from this equation to (log a)/(log b) = (log b)/(log a)?

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