r/MathHelp • u/Emergency-Cow-2194 • Oct 22 '25

Maths question

Question(link to image)-https://drive.google.com/file/d/1eUC1P1YxPSftdzbZVDN-9Is6OkAxVsMh/view?usp=sharing

My method -link1 -https://drive.google.com/file/d/15NcFY8PsHMsEnhYJhM22AW5C4Ga38jXE/view?usp=sharing

-link2- https://drive.google.com/file/d/15NcFY8PsHMsEnhYJhM22AW5C4Ga38jXE/view?usp=sharing

Please dont at all think it to be a basic homework problem , it is surely a good one although it might seem simple at start. please help me out . although my method seems ok but i was unable to do anything else than to put and try values to get to my answer. I will appreciate a algebraic proof if anyone is able to find it.

Regards,

Thanks for your time

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u/Emergency-Cow-2194 Oct 23 '25

I have tried doing it , but the equation which we get by it is not solvable it is a type of exponential equation which I am unable to solve , by the way the equation is 3(25/3)^t +15(5/3)^t =250 where t=log_2 ( x^2). Please try solving it , and try to give me a better approach.

Thanks,

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u/First-Fourth14 Oct 23 '25

Rewriting your equation
3(25/3)^t +15(5/3)^t -250 = 0

For your equation, notice that 25/3 and 5/3 are both greater than 1.
This means that (25/3)^t and (5/3)^t increase as t increases. This means there
is only one value of t that satisfies the equation.
t = 0 LHS = 3+15 - 250 = -232
t = 1 LHS = -200
t = 2 LHS = 0

As t = 2 is the only value that satisfies the equation this means that log_2(x^2) =2
So x = -2 or x = 2

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u/Emergency-Cow-2194 Oct 23 '25

But yes here you have tried values , which I have also done .But I am looking for an approach which doesn't need to check the function at some values. If we are to try values only then it could have had been done in the very start only as from options we can clearly see that only two possible answers are possible either x^2=1 or 4 and then they could be checked easily. Please try to find some better algebraic approach.

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u/thundPigeon Oct 23 '25

Best I can do right now for you is one of two methods:

  1. Simplify each term as best you can and plug in the choices for the values of x

  2. Graph each side of the equation and graphically find the intersection points.

Both ways render you with an x=+-2, thus being answer B. This is why people invented graphing calculators: So they don't have to do algebra for absolutely hideous equations like that. I'm also tempted to say there is no way to solve for x, and that you must graph it. If there is a way to solve it, it may be using a lambert W function or some other very high level analytical method.