Proving Unique Solution for ab = ba Using Logarithms | Homework Equations

[songoku]

Homework Statement

Positive integers a and b, where a < b, satisfy the equation :
a^b = b^a

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

Homework Equations

logarithm

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

 

[rock.freak667]

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

 

[songoku]

Hi rock.freak667

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

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

 

[turin]

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

 

[icystrike]

Since a and b are positive integer,

we can conclude a,b>0 without generalisation.

First by taking log base a,

$$<br /> b=a log_{a}b$$

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

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 (=

 

[songoku]

Hi turin and icystrike

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

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?

substituting to initial equation,

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

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

$$a=0$$ 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 (=

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=a^a, how to deduce that a = 2?

Thanks a lot

 

[Mark44]

Since a and b are positive integer,

we can conclude a,b>0 without generalisation.

First by taking log base a,

$$<br /> b=a log_{a}b$$

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

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.

substituting to initial equation,

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

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 log_ab.

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

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

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

Assuming that the base is larger than 1, any log curve is constantly increasing. IOW, if x₁ < x₂, then log_ax₁ < log_ax₂.

$$b=a^{a}$$

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

 

[turin]

Assuming that the base is larger than 1, ...

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

 

[Mark44]

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

 

[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

 

[Mark44]

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.

$$((lg a)/(lg b))=((lg b)/(lg a))$$

How do you get the equation above? We know that a^b = b^a, 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)?

$$(lg a)^{2}-(lg b)^{2}=0$$
$$(lg a-lg b)(lg a+lg b)=0$$
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