Maths problem you may find easy

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"Given that 64^y=2^3y-1, find the value of y."

I can do this with a calculator, but it is a non-calculator paper...

Anyone know how to go about this mentally?

Edit:

As soon as I posted I realised how... d'oh...

Anyway, I won't put up the answer or working so anyone who wants to can do it

Ok, i think i have it . . .

64^y = 2^(3y-1), y = ? 2^6 = 64

2^(6y) = 2^(3y-1) -----> 6y = 3y-1 -----> 6y - 3y = -1 -----> 3y = -1 -----> y = -1/3

That is what I got when I thought I'd realised, but its also what I got before I realised I just forgot that was what I originally did :-S

But now put it back into the original equation and it doesn't seem to work and I don't get why not.

3.81469727 × 10-6 != 0.25

I think we've both done something fatally wrong

I tried putting it into online equation solvers and COMPUTER ASPLODE! is all I get back....

You probably just entered it wrong.

64^(-1/3) = 2^(3(-1/3) - 1)
2^(-6/3) = 2^(-2)
2^(-2) = 2^(-2)
0.25 = 0.25


Now, when I first read your question I thought it's
64^y = 2^(3y) - 1

Does anyone know if it is possible to solve?
1 = 2^3y - 2^2(3y) doesn't seem plausible.

I did indeed... I trust my computer to get the correct anwswer more than me...

Buuutt. Whereas -1/3 was in my head (-1/3) google reads it as 64 to the minus one and then divide by 3 as opposed to what I see in my head as being 1 over three confused by the slash for divide... Even though it means the same thing. If this makes sense?

compare 8^(2k) with 8^k

how small can 8^(2k) - 8^k get when k -> -infinity this difference goes to 0 from above so it can never be equal to -1.

hence the equation 64^y = 2^(3y) - 1 has no solutions. QED.

ruveyn

Thats why i grouped up the exponents, it wouldnt have made sense otherwise. There is probably a way to solve the other version without, but its likely a higher form of math im not familiar with. >< So, as far as i can tell, 64^y = 2^(3y-1) was the original setup/intent of the problem. Let me see here . . . if not:

2^(6y) = 2^(3y) - 1 . . . > 2^(6y) - 2^(3y) = -1 . . . > (2^(2y) - 2^y)^3 = -1 . . . > 2^(2y) - 2^y = (-1)^1/3

> 2^(2y) - 2^y = -1

The answer to this problem doesnt exist because 2^(2y) will be > 2^y, so a negative answer is impossible.

Exponents
Parentheses
Multiplication
Division
Addition
Subtraction

The only thing I learned in 16 years of math...

Thanks, I got so absorbed in solving it I forgot about testing it with limit.



I don't think there's a solution either. But I'm not sure how to prove it.

The way I did it:
2^3y - 2^2(3y) = 1
8^y - 8^2y = 1

When y = 0, 8^y - 8^2y = 0
When y = -0.1, 8^y - 8^2y = 0.15249844096978839292270839716046
When y = -0.2, 8^y - 8^2y = 0.22447867373838506011886597687495
When y = -0.4, 8^y - 8^2y = 0.24581071083426230927472753357657
When y = -0.5, 8^y - 8^2y = 0.22855339059327376220042218105242
When y = -0.6, 8^y - 8^2y = 0.20470534432595286048878161349263

So, it's possible 8^y > 8^2y, it's just not possible for the difference to be 1.

I think the correct way is to differentiate 8^y - 8^2y to find all the points where slope = 0. And then use limit to test each case. The problem is I don't know how to do that. By inference from the table, I guess the peak should be around y = -0.4.



Me neither, that's why I'm curious if there's a way to handle it.

The differential of 8^y-8^2y is 3 ln(2) 8^y - 6 ln(2) 8^2y, where ln is the natural logarithm. Putting this equal to zero we have:

3 ln (2) 8^y - 6 ln (2) 8^2y = 0.

Simplifying and rearranging:

8^y = 1/2 or

2^3y = 2^(-1).

Equating the indices:

3y = -1.

Therefore the function 8^y - 8^2y peaks at:

y = -1/3.

Plugging this into the original equation gives:

8^(-1/3)-8^(-2/3) = 1/2-1/4=1/4.

Therefore 8^y - 8^2y = 1 has no solutions.



If such an equation did have solutions, then there is no way to solve it algebraically. It can however be solved with root finding algorithms. I can give you an example of such an algorithm if you like but its tedious to do such a calculation by hand. It's best to use a computer with numerical analysis software that can perform the calculation automatically.

Thanks for doing the hard work.

Does that mean the only way is trail and error, albeit in a smarter way?

I'm not sure if you could call it trial and error. First note that root finding algorithms solve equations of the form f(x) = 0. So before using one, everything must be taken over to the left hand side and this becomes your function f(x). The simplest root finding algorithm is the bisection method. To use the bisection method, apply the following algorithm:

1. First choose an error tolerance E. For example if you want to calculate the solution to within 4 decimal places, take E = 0.0001.

2. Now you need an interval [a, b] which contains the solution. This is why you have to solve for the equation f(x) = 0, because then you can easily see that you have such an interval if f(a) < 0 and f(b) > 0 or visa versa. You can obtain an initial interval by plotting a graph or creating a table of the function and taking the interval from there.

3. Now take c = (a+b)/2 and test it by plugging it in to your function.

4. If |f(c)| > E, then check f(c) is positive or negative. Say for instance that with your initial interval f(a) < 0 and f(b) > 0. Then if f(c) < 0, put a = c otherwise put b = c and repeat from step 3 with your new interval. If |f(c)| < E then stop and take c as your solution.

When applying the above procedure you will find that the number c gets closer to the solution the more times steps 3 and 4 get repeated. There are other root finding algorithms that get to the answer faster with less iterations but the above one is one of the simplest. A computer can be programmed to implement such an algorithm automatically but I think that it can done with a graphing calculator as well.

There are iterative algorithms for approximating the zeroes of such equations to any desired degree of precision.

To see such algorithms pick up any text book on numerical methods.

ruveyn

Everyone, thanks again for the clear explanations.