X is - ve 4 sure .. rest is easily find

bro wasted so much time doing the 2^0 when its common knowledge that it = 1

X = 2, right?

3 trivial solutions: 2, and 2 with 1/3 or 2/3 of a full rotation. ie 2*e^2ipi/3, and 2*e^4ipi/3.

(2^x)^3+2^x=3^3+3 f(z)=z^3+z is a bijective function R-->R so 2^x=3, x=log3/log2

The trick involves recognizing that 30 is the sum of a number and its cube. However, if you can do that, you can just solve the original equation by inspection once you have made the substitution y = 2^x. I suppose an advantage of this factoring approach is finding other real roots of the cubic, if they exist.

If you're already dealing with complex numbers why not apply radiation formula?!?! Saves so much time... x^3 = 8 => x^3 = 8e^i0 => x = 2e^i2kpi/3, let k = {0, 1, 2} then S = {2e^i0 (or simply 2); 2e^i(2pi/3); 2e^i(4pi/3)}

This is insane...I'm impressed...and would you consider reducing the volume of the music?

= 10

8^x+2^x=2^(3x)+2^x=6^3+6, f(z)=z^3+z is a monotone function, so 2^x=6 x=log6/log2 which is same as 1+log3/log2

x=0.5 is a trivial solution. x^6 cancels out so this is a polynomial of degree 5. The other 4 solutions must also have x and x-1 of equal magnitude and have equal complex angle taken to the 6th power. Check 2pi/6 as trivial angle in unit circle, since e^2pi*i/6-1=e^2pi*i/3. This gives 1/2+sqrt(3)i/2 and its complex conjugate as a solution. Now we can find the remaining quadratic of 6x^2-6x+2=0 which gives the other 2 solutions.

Nice one

C

Ohh 😮 i've got so used that we solve everything in R, that i just solved it in R and forgot to check the C

After 4th row should be used log2

Nice

0.5 in the real world is so obvious. Then you fill in X=a + bi and solve it.

0.5^6=(-0.5)^6 Before watching, just my guess

9⁹

√2i

Nice

Very easy, but your solution is very confusing and hard.

As an Indian eight grade student I was able to solve as soon as I saw it 😂

This could've been a 1 minute video.

this might be a stupid question but why wouldn't you just multiply and add them

How was I able to follow that with no more understanding of higher math than: ''What you do to one side you do to the other" ?

I have seen many videos about finding complex roots. However, I have not seen one using De Moivre's theorem except the videos specifically descibing this theorem. This video's will probably be cut in half if the theorem was used.

As others have said, no need for logs, just do it by inspection. 3x2^2x = 24; 2^2x = 8 = 2^3; 2x=3; x=3/2

You can also just use the sum of a goemetric progression formula: (a(r^n-1))/(r-1) where a = 9, r = 9, n = 5. (9(9^5-1))/8 = 66429

4^x = 8; 4=2^2; 2^(2*x) = 8 --> also, 2^3 = 8 --> 2*x = 3 --> x=3/2; [gets the calculator out] x = 1.5 (I'm a maths dad trying to keep up with the kids but does this answer check out! I forgot about logs!)

12

3*2^2X=24, 2X=3, X=3/2

3^m > 65. test m=4. 3^m =81. good start. 2^m=16. m=4. rest my case

0.2 approx mind calculation

i used logarithms immediately and used a calculator to get the exact values

1) a=25;b=24 2) a=7;b=0

Isn't log (to the base of 6) of 6 just 1? Asking 4 a friend ;-)

18 =6^1.613....

Taking (9^x) as (U) and then simplifying the equation is easier, actually!

Log 4

1

So a^2 - b^2 == 49 == 7^2. Fortunately I know a trick that there is a 7, 24, 25 right triangle. Next video !

X=20

(3x)^2 = 9x^2

Of course, this perform 0 as default With +-7 as pairing I mean a=+-7, b=0

-3

This is trivial. Divide by 3 and then take the logarithm to base 9.

You never mentioned that a and b have to be whole numbers, so there's infinitely many solutions. a=sqrt(49+k) and b=sqrt(k) where k is a real number bigger than 0 Checking: 2a²-2b²=98 / div by 2 (sqrt(49+k))²-(sqrt(k))²=49 abs(49+k)-abs(k)=49 We can ignore the absolute value (because k is bigger than 0 - therefore 49+k and k will always be bigger than 0) 49+k-k=49 49=49 QED Example: k=2 => a=sqrt(51) and b=sqrt(2) 2(sqrt(51))²-2(sqrt(2))²=98 2*51-2*2=98 102-4=98 98=98