Exponential Equations With Prime Number Multiplication

[MayaMaya121]

Hi,
I wanted to ask about a solution for an exponential system of equations. If I have the following equation (it's one equation out of the two):
[MATH]5^{2x}\cdot 3^y=675[/MATH]Can I say that x=1 and y=3? (Since [MATH]675=5^2\cdot 3^3[/MATH])

Thank you for helping.

 

[Dr.Peterson]

Only if you are told that x and y are integers.

 

[topsquark]

Hi,
I wanted to ask about a solution for an exponential system of equations. If I have the following equation (it's one equation out of the two):
[MATH]5^{2x}\cdot 3^y=675[/MATH]Can I say that x=1 and y=3? (Since [MATH]675=5^2\cdot 3^3[/MATH])

Thank you for helping.

Hint: Pick a base logarithm and take the log of both sides. (I usually use base "e.") What do you get on the LHS?

Give it a go and post back here with your work if you get stuck.

-Dan

 

[Dr.Peterson]

I wanted to ask about a solution for an exponential system of equations. If I have the following equation (it's one equation out of the two):
[MATH]5^{2x}\cdot 3^y=675[/MATH]Can I say that x=1 and y=3? (Since [MATH]675=5^2\cdot 3^3[/MATH])

If you tell us the whole problem, including conditions on the variables and the other equation, we'll have a much better chance of giving you useful advice.

Of course, if x and y were required to be integers, then you wouldn't need the second equation, so it's almost certain that isn't true. But the other equation may suggest a much easier method of solution than we expect. And, of course, it's possible that the solution is in fact (1,3), and the author of the problem unwisely made the solution too easy to find by inspection.

 

[MayaMaya121]

Hint: Pick a base logarithm and take the log of both sides. (I usually use base "e.") What do you get on the LHS?

We still haven't learned in class about log base "e". But I see another possible solution using another base.

If you tell us the whole problem, including conditions on the variables and the other equation, we'll have a much better chance of giving you useful advice.

There are no conditions on the variables. The other equation is:
[MATH]\log _{\sqrt{2}}\left(x+y\right)=4[/MATH]I solved the equation system, but I was curious about the first one I wrote...

 

[Dr.Peterson]

I think my guess is right: You can't assume x and y are integers, but it turns out that they are because the author chose to have a "nice" solution, namely (1,3).

This can be solved without making the assumption; I hope that's what you did: Solve the second equation for y, which is easy, then substitute in the first equation and manipulate to get an equation that is easily solved. I'd be interested to see if your method is the same as mine.

If the equation were changed just a little, it would at least require logs, and might become impossible to solve exactly. But as it is, you could just obtain (1,3) as a solution by inspection, and then confirm that it must be the only solution because of the behavior of the functions involved.

 

[MayaMaya121]

This can be solved without making the assumption; I hope that's what you did: Solve the second equation for y, which is easy, then substitute in the first equation and manipulate to get an equation that is easily solved. I'd be interested to see if your method is the same as mine.

Yes, that's what I did.

If the equation were changed just a little, it would at least require logs, and might become impossible to solve exactly. But as it is, you could just obtain (1,3) as a solution by inspection, and then confirm that it must be the only solution because of the behavior of the functions involved.

Ok.

Thank you all for helping!

 

[JeffM]

Your second equation implicitly did impose a condition. You can think of a system of equations as each equation imposing a condition, namely that that equation must be true at every valid solution.

Lastly, e is an irrational number like pi. And e and the logarithm using e as a base are fundamental in much of calculus. Therefore, many people familiar with calculus tend to use that logarithm almost exclusively, and some call it the natural logarithm. But, like pi, e is not exactly expressible in a finite number of digits. The only reason to learn about e and the natural logarithm in algebra is to realize that (a) e has all the properties of any number > 1, and (b) the logarithm using e as its base works algebraically just like a logarithm to any other base > 1. That way, if you get to calculus and start working with e and the natural logarithm, they will be somewhat familiar, and you can concentrate your attention on what makes them different and so important in calculus. But, for an algebra student, e and the natural logarithm are just another number and logarithm, nothing strange about them at all.