Відео

Took me less than 15 seconds to find all the solutions

lowest nonzero term of x is z^2, giving x=0 with multiplicity 2 leaves x^4 - 5x^2 + 6 = 0 let z be x^2, apply any quadratic solving method attains z = 2 or 3 results in 0 , +/- sqrt(2 or 3) as all values

Its wrong, the others roots are 1±i [square root of 11] And 3

3ᵏ + 3ᵏ = 150 2*3ᵏ = 2*3*5² 3ᵏ = 3*5² k = 1 + 2*log₃(5) = 1 + 2*log(5)/log(3) ≈ 3.9299470... If you have a simple 4-function calculator and still know log(2) and log(3) to 5 significant figures: k = 1 + 2*(1 - log(2))/log(3) ≈ 1 + 2*(1 - .30103)/.47712 ≈ 3.929954728... ≈ 3.9300 { 5 significant figures }

7ˣ + 7ˣ = 490 2*7ˣ = 2*5*7² 7ˣ = 5*7² x = log(5)/log(7) + 2

Good work my fellow Mathematician ❤ keep up I have subscribed

I solved it as X=1+log15/9 Is it correct or not?

K= 3.93

X= 2.8271

Well explained

Mistake. You cannot move 3 before log as you wrote

plus astucuex en 3 lignes : 2.7^x = 7^2. 10 --> 7^(x-2) = 5 --> x-2=ln(5/7) --> x=2+ln(5/7)

?????????????????😢😢😢😢 Não ⛔ vale isso!!!! Brasil Novembro de 2024.

That could have been solved in seconds?!

@5.52 how 5 comes?

Решение в три действия, и занимает 1 минуту, не позорьтесь 1 Действие, 7^х*(1+1)=490, 2*7^х=490 2 действие, делить на 2 7^х =245 3 действие логарифмировать по основанию 7 х=log(7)245 Всё решение Можно еще 245=5*49, тогда ответ х=2+log(7)5

x = 9: 9^9 = 9^9 = solution, (no calculation necessary). x = 1: 9^1 = 9, 1^9 = 1, no solution x = 0: 9^0= 1, 0^9 = 0, no solution

Sometimes u wrote it is a polynomial equation and sometimes exponential one. Polynomial equation The variables are the bases. Exponential equitation The variables are exponents.

x^2=2x+3 , x^2-2x-3=0 , x=(2+/-V(4+12))/2 , x=(2+/-4)/2 , x= 1+2 , 1-2 , x= 3 , -1 , test , 7^9/7^6=7^3 , 7^3=343 , 7/(1/49)=343 , OK , solu , x= 3 , -1 ,

If we do the thing like this: 9^(x-2) = 54; log_9(9^(x^2)) = log9(54); (x-2)log9(9) = log9(54); x-2= log9(54); x = log9(54) -2; is this a correct solution?

Factors of -6 divided by factors of 1 are possible zeroes. 1 is the easiest to use. Try it. Synthetic division shows the polynomial is divisible by (x-1), so x=1 is a factor. What's left is x^2 - 5x + 6 The factors of 6 that add to -5 are -2 and -3. (x-2)(x-3)=0 x = 2 and x=3 The zeroes/solutions are x=1, 2, and 3. BTW, it's not an exponential equation. It's a polynomial equation. Exponential equations involve exponents that are variables.

Good work done

Why people donot use better dark pen??

x=3+log 13 base 7

8 mins ???? mental arithmetic 15 secs

=13/8=1.625

Well explained

1:29 easy enough to solve here, all the following work is not needed. sqrt(9) = 3 so the answer = 4/2 = 2

Before doing any stretched calculations. By inspection only looking at the equation you can see that if the 3 into brackets could be reduced to 2 then everything would be over. To do that the x should be -1 or -5.

(k+1)(k-2)(k^2+k-1)=0 ,

X+3 = ±2i X = ±2i-3 X+3 = ±2 X=-5 or x=-1 We know that equation will have 4 roots and we can easily see those roots so we can directly equate like i did above

by faktoring , (x-9)(x^2+8x+72)=0 , x=9 , x=(-8+/-V(64-288))/2 , x= -4+i*V56 , -4-i*V56 ,

Its doubled squared in the photo

I like to add a way to evaluate 3^9 3^9=3^4*3^4*3 =81*81*3 =6561*3=19683 8 1 * 8 1 --------------------------- 64 1 1 6 ---------------------------- 6 5 61 Or 81*81=81^2=(80+1)^2 =6400 +1+2 80*1 =6401+160=6561

💯

I just looked at the thumbnail, thought about it for like 30 seconds and got the answer

This was too easy

There are probably much more complicated ways to solve this problem 🤣 1. X != 0, 2. 5*5 = 25 3. x*x=x2 4. multiply the equation with 5 and divide by x (because it has to be != 0, still an assumption) 5. 5³ / x³ = 1 ==> x³ = 5³ ==> x = 5, because 5³ > 0 . That's it, right?

7^x + 7^x =490 > 7^x =245 x =log 245 (base 7) = log (7^2*5)(base 7) = log 7^2(base 7) + log 5(base7) = 2log 7(base 7) + log 5(base 7) = 2*1 + log 5(base7) = 2 + log 5(base 7)

In the last line you wrote 2+ log 5/log 7 And in margin wrote log a /log b = log a (base b) But in answer you did not write 2+ log 5(base 7)

EXPONENCIAL

Simpler solution: Sqr root both side gives (x+3)^2=+/-2^2 Sqr root again gives (x+3)= +/-2 or +/-2i x = -1 or -5 or -3+2i or -3-2i 🤷🏻

At 2:30 you make and error saying that it = a^2-ab+b^2. It is, in fact = A^2-2ab-b^2. You use the correct expansion in the square brackets after that though.

Damn bruh

My way: 3ˣ·3ˣ·3ˣ = 30 => 27ˣ = 30 => x = ln(30)/ln(27) or x = log₃(30)/3 🙂

7^x+7^x 7o^70= 70*70=70^2= 7^20]=[ 2*7^x =7^2x= 7^2o 2x= 2ox=1o ACADEMIC Marcelius Martirosianas

Use Euler's equation, x_j=2*cos(jπ/2)+2*i*sin(jπ/2)-3, j=0,1,2,3, to obtain four roots, where i=√(-1)

3

That is much too circumstantial. You should cancel as much as possible in the very beginning: 7^x + 7^x = 490 => 2*7^2 = 2*49*5 => 7^x = 7^2 * 5 =>7^(x-2) = 5 => ln(7^(x-2)) = ln(5) => (x-2)*ln(7) = ln(5) => (x-2) = ln(5)/ln(7) => x = ln(5)/ln(7) + 2. Done.

Thank you well explained ❤