Weekly Math Question

I'm thinking of posting a new math problem every week here if you guys like the idea of it. Anyway, here is the first Math Problem, I'll probably post the solution to it on Saturday:

Given the equation y = (x^n) + x^(n-1)
Solve this expression for n.

Pro Question: If the above equation is to easy for you, try this:

Given the equation f(x) = (a^g(x)) * (ln(a) + 1)
Solve this expression for a.

Anyone is welcome to try and solve these questions, but please don't cheat by using high tech software. Good luck everyone!

If only it was odd-numbered (I'd have the answer for you in a jiffy!).

@nekkoteki I believe you are just forgetting about logarithms, I'm not sure if you learned it yet in your schooling, if not then it might be unfair for you to do this, and you will be giving yourself a headache for nothing :P. Note in the advanced question he uses "ln" which is the natural logarithm with base e, euler's number. If you know about logs then it is trivial to remove the exponent.

Also its better to factor out the x^n, personally I think the end result is cleaner.

@kyanite111 said in #4:

@nekkoteki I believe you are just forgetting about logarithms, I'm not sure if you learned it yet in your schooling, if not then it might be unfair for you to do this, and you will be giving yourself a headache for nothing :P. Note in the advanced question he uses "ln" which is the natural logarithm with base e, euler's number. If you know about logs then it is trivial to remove the exponent.

Also its better to factor out the x^n, personally I think the end result is cleaner.

Factoring out x^n gives you y = (x^n)*(1+x^(-1)), leaving you with negative exponents, which imo isn't the cleanest way of doing this problem.

Google and ChatGPT has entered the chat

Has anyone tried the pro question so far btw?

Alright, last day! Tomorrow I will reveal the answer and the next question. Do you guys like the concept of this so far?

Alright, time to reveal the solution to our very first questions:

Let’s take our equation

y = x^n + x^(n-1)

Factoring out x^(n-1) gives us

y = x^(n-1) * (x+1)

Dividing both sides by (x+1):

y/(x+1) = x^(n-1)

Next, we take the logarithm to the base x and rewrite it with the natural logarithm (log_x(z) means the logarithms to the base x of z):

n-1 = log_x(y/(x+1))

n-1 = ln(y/(x+1))/ln(x)

Finally, rewrite ln(y/(x+1)) by using basic “log”- rules, add one to both sides and we get

n = (ln(y) – ln(x+1))/ln(x)

As we see, because the denominator shouldn’t get zero, we have to include that ln(x) mustn’t be equal to 0, which means x can’t be 1 (test it yourself). So, the final result is:

n = (ln(y) – ln(x+1))/ln(x); (x ≠ 1)

However, anyone who got

n = log_x(y/(x+1)) +1

also got the correct solution. The above solution just shows how I would have solved this problem.

For our pro question, I’ll have to introduce a new function that many of you probably haven’t heard of: The Lambert W Function. You can go to Wikipedia for details, but it pretty much does this: If you have f(x)* e^f(x) and apply the Lambert W Function on this, you will get f(x):

W(f(x)*e^f(x)) = f(x)

So how does that help us with our problem? First of all, let’s set f(x) = y and g(x) = g for simplicity. In the end, we can substitute f(x) and g(x) back into our solution. So now we have

y = (a^g)*(ln(a)+1)

As we all probably know, e^ln(x) = x, hence x = e^(ln(x)). Applying this fact to a^g gives the equivalent term e^(ln(a^g)). Using a basic property of the logarithm function, we can rewrite this as e^(g*ln(a)). Putting that in our equation, we get

y = (e^(gln(a)))(ln(a)+1)

By multiplying both sides with e^g, we get

y*(e^g) = (e^(gln(a)))(e^g)*(ln(a)+1)

Applying another logarithm property, we get:

y*(e^g) = (e^(gln(a)+g))(ln(a)+1)

y*(e^g) = (e^(g(ln(a)+1)))*(ln(a)+1)

Multiplying by g, we get:

gy*(e^g) = g(ln(a)+1) *e^(g(ln(a)+1))

Now we can apply the Lambert W Function on both sides (yes, this is legal), and we get

W(gy*(e^g)) = g(ln(a)+1)

Now we can simply solve for a:

W(gy*(e^g)) = g(ln(a)+1)

W(gy*(e^g))/g = ln(a)+1

(W(gy*(e^g))/g)-1 = ln(a)

a = e^((W(gy*(e^g))/g)-1)

Substituting f(x) and g(x) back, we get our final answer:

a = e^((W(f(x)g(x)(e^g(x)))/g(x))-1); (g(x)≠0)

Ok, here's the question for the next week. The answer will again be revealed on Saturday next week.

Given that x + y + z is divisible by 3 (this means you can divide x + y + z by 3 without getting a remainder). Prove that (x^3) + (y^3) + (z^3) is also divisible by 3.

Pro Question: For anyone who thinks he's better, try this.

Prove that if x + y + z is divisible by a number, (x^3) + (y^3) + (z^3) is also divisible by that same number.

Have fun!