It is not right to substitute 392=343+49 without providing a method of finding. a^3+a^2=392 but 392=2*2*2*7*7 a^2*(a+1)=392 You can find the solution in these divisors and their combinations. You easily find a=7 as first answer transform the initial equation to a third-degree equation 1*a^3+1*a^2-392=0 and compare coefficients to a third-degree equation with one known answer: (a-7)*(A*a^2+B*a+C)=A*a^3+(B-7*A)*a^2+(C-7*B)*a-7*C=0 hence A=1 B-7*A)=1 B=8 -7*C=-392 C=56 Find complex results; a=(-B+-SQRT(B^2-4*A*C))/(2*A)=-4+_i*2*SQRT(10) Your complex results are wrong!

3^21

The trick is you figure out that one of the roots are 7. From there its simple polynomial devision and second order equation solution. For me, I did trial and error. I said 10^3 is 1000 which is way more than 392. I then tried 8^3 and it was again more than 392. I tried 7^3 and bam, that’s one of the roots. Now if none of the roots are whole numbers, good luck with your solution. Try solving for x^3 + x^2 = 400 and let me know how it goes.

I thought log a/log b = log (a-b) or doesn’t that apply here?

Your answer to the cubic is wrong. Messed up with the radical.

X^X^6=144 X=±Surd[12,6]

10460353203=3^21

For the first problem, I simply said x = (log90)/(log45) = 1.1821.

My solution is simply x = (log15)/(log3) = 2.465, which is different from yours. Please check your answer.

Yeah, that's not right. log(5)/log(3) is 1.464974 and the answer should be 2.464974.

Please check your solution for 3^x=15. 3^2.4851=15.335 while simple solution x=log(15)/log(3)=2.465 and 3^2.465=15.000436

ln(90)/ln(45)

a³+a² = 7³+7² a³-7³+a²-7² = 0 (a-7)(a²+7a+49)+(a-7)(a+7) = 0 (a-7)(a²+7a+49+a+7) = 0 a = 7 a²+8a+56 = 0 a²+8a+16 = -40 (a+4)² = -40 a = -4±2i√10

Note that zero is not a solution

X^x = x^2 Sqrt(x^x)= sqrt(x^2) (x^x)^(1/2) = x X^(x/2) = x = x^1 X/2 = 1 X = 2

Before finding the actual solutions to a quadratic you can stop if delta of the expression is lt 0 if you do not want conplex values

But fo x=2 2^(3*2)+2^2=2^6+2^2=64+4=68 since x=2 is not a solution. The solution x=1 is easy to guess immediately

Decimal 10 is binary 1010, which is 2^3 plus 2^1. Therefore by simple inspection, x=1. No "genius" level mathematical skills required. Didn't need an 8 minute video, either.

Sir without doing all this just split 10 as 2 to the power 3 plus 2 to the power 1 then compare both the sides

2 · 1/2 · 1/2 · 1/2 = 1/2 · 1/2 = 1/4.

As you did from the 4 min mark, I also like to avoid long multiplication. This is my effort: 2^18=2^10×2^8=1024×256 =1000×256 + 24×256 =256,000 + 6x1024 =256,000 + 6,000 + 6×24 =262,000 + 12×12 =262,144 2^18 - 1 = 262,143. Thanks for posting the maths.

=) The long way eh?

... ln(x^x)=ln(x^2) ==>xlnx=2lnx ==> x=2 or 1

The same base (X) is present on both sides of the equation, then X=2 is an immediate solution. And, of course, since 1 to any power is equal to 1, 1 is the other solution.

I agree with -8.

Daft question

Let me guess you saw somebody else do this so you thought you would just remake it with your crappy PowerPoint presentation and drag it out for the algorithm

Is it wrong of me to be able to identify 1 and 2 as solutions in the first 3 seconds? 😏

Isn't the answer equal to: -8 * (9^900)

Nonsense, not "IMPOSSIBLE" at all. Not even hard. Not even _moderately_ hard. 7^x=70 ⇒ x=1+ln(10)/ln(7)≈2.18329466245494. That took me 7 seconds to compute. Why did it take you 15 minutes? And why did you think I couldn't compute it? 🙄

This is incorrect. By exponent rules, a^m^n =a^mn. Your answer works because you changed the base from 2 to 3 (random coincidence for this problem). The problem can be simplified to 2^12x = 2^9. x=3/4 or .75. Check with a calculator.

512=2^9. 3^4=81. 2^81^x=2^9, 81^x=9. 81=9^2 so 9^2x=9. sqrt(9^2x)=+/-9^x=sqrt(9)= +/-3. x=1/2.

You do know, Math Person, that, 0 does not equal 0. It can but not all the time. The key word there is-tjme.

What about simply 7^x = 70 log7(7^x) = log7(70) x = log7(70) x = ln(70)/ln(7) x ~= 2.18329 Doesn’t seem to be so impossible tbh

You cant cancel things out. There is no operation of canceling in maths ;)

2^10 = kibi = 1024, 2^20 = mibi = 1024 * 1024, 2^30 = gibi. So 512 = kibi/2 = 2^9.

This is the same moron who thought he could prove that 0/0 = 2 in another video.

"An IMPOSSIBLE Exponential Equation"?? It's practically *trivial*. Go away.

wtf is the point if calculator being used anyway just type in log ₇ 70 at step 1

What the hell, the problem is so easy and you even overcomplicated it 7^x = 70 xlog7 = log70 x = log70 / log7 x = 2.1832946.....

The mouth and swallowing noises provide some nice ambiance

x = log4(log3(log2(512)))

'Ministry Of Love' logic from 1984.

8^(1/3) = 2 2^3 = 8 2^5 = 32 remember that 2 = 8^(1/3) therefore: 2^5 = 32 8^(5/3) = 32 2 elevated 5 times is the same as 8 elevated a 1/3, 5 times and also, 5/3 is the same as 1 2/3 as someone else said it

Man...my is teacher was teached me wrong

Its been proven that pi is 3.141..., check it yourself with a rope and ruler

Nonsense. What you’ve only shown is that 0a = 0b . Yes, that holds, but it doesn’t imply that a = b.

they forgot to take the abs of them when puling out of the root (4:20)

That is not how you prove pi = 3 ! because with the same logic you could use any value to make it = 3 or what have you. Proof in maths are a lot more demading and the solutions need to be unique which is not the case as you can obviously see and determine.

You're = you are. Your*