1. ## exponent/decimal equation

My daughter and I are having trouble figuring out how to do a certain type of problem... The trick is to figure out what the exponents are.

So, here's an example:

1.5^x + 8^y = 633

I can't even begin to figure this out...

Any help or explanation would be appreciated!

Math-impaired mom

2. Originally Posted by dphock
My daughter and I are having trouble figuring out how to do a certain type of problem... The trick is to figure out what the exponents are.

So, here's an example:

1.5^x + 8^y = 633

I can't even begin to figure this out...

Any help or explanation would be appreciated!

Math-impaired mom
this question seems a bit too hard for Middle school math. i can't really figure out a nice way to find a general solution without calculus.

what guide lines were given? are we allowed to say pick the value of one exponent and then find the value of the other that makes the equation true, etc?

3. ## Re:

Thanks!!

4. Originally Posted by qbkr21
Thanks!!
Nice solution. still seems too complicated for Middle school though. although, i have no idea what Middle school is like, don't know much about the American school system

5. I just had to reply to this one since my daughter got this as well. There were 4 problems with one of them matching the one in the previous email:
1.5^a + 8^b = 633
2.6^c + 2^d = 88
3.2^e + 3^f = 499
4.1^g + 9^h = 3^i

Where a-i are all unique numbers from 1-9.

On the page, they didn't have any space betweeen "1." and 5 and the were written randomly across the page with many fishes in between

The real problems were:
1) 5^a + 8^b = 633
2) 6^c + 2^d = 88
3) 2^e + 3^f = 499
4) 1^g + 9^h = 3^i

We spent about 1 hour with 3 adult helping out to finally figure this one out. Just thought someone else might run into this.

6. I'm pretty sure that's copied wrong.

2) 6^c + 2^d = 88

should actually be

2) 6^c + 2^d = 68

If all of the variables are integers, then the solution of 88 is impossible.

I need to read the rules. Can I post the answers here?

7. Hi Denmark,

I'd appreciate any insight. I believe that the problem should be 88. The solution that my daughter came up with was:

6^3 - 2^7 = 88
216 - 128 = 88

By the way, this is for my 4th grade daughter who was stuck on this. It is in an optional math packet that she is working on.

In the end, neither of us could come up with a solution to this problem. She had a solution for each of the 4 problems, but the requirement was that each problem only use one of each of the numbers 0,1,2,3,4,5,6,7,8,9.

I ended up writing a short perl program last night to try all combinations and I came up with the same answers:
1**z + 9**x = 3**y
X=2 Y=1
X=4 Y=2
X=6 Y=3
X=8 Y=4
5**x + 8**y = 633
X=4 Y=1
6**x - 2**y = 88
X=3 Y=7
2**x + 3**y = 499
X=8 Y=5

There are many solutions to 1^z + 9^x = 3^y. However, they all are reusing one of the integers that was used in the other 3 problems. She has the worksheet at school now to ask the teacher about it.

8. Okay.

After several ridiculous errors, here is my proposal for what the homework should have been:

1) 5^a + 8^b = 633
2) 6^c + 2^d = 548
3) 2^e + 3^f = 499
4) 1^g x 9^h = 3^i

How 548 turned into 88, I don't know. Nor how the operator for problem 4 turned into addition from multiplication.

Regardless, the above is very similar to the given problem, and satisfies the requirement that all of the variables be unique integers between 1 and 9 inclusive.

a = 4
b = 1
c = 2
d = 9
e = 8
f = 5
g = 7
h = 3
i = 6

1) 5^4 + 8^1 = 625+8 = 633
2) 6^2 + 2^9 = 36+512 = 548
3) 2^8 + 3^5 = 256+243 = 499
4) 1^7 x 9^3 = 3^6 = 1x729 = 729

9. Hi Denmark,

Thanks for the help. I finally got a chance to talk to my daughter's teacher and there was a flaw in the problem. The solution page had an answer that re-used the number 1. I shared your examples with my daughter and she found them very interesting.

Thanks again.