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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 ❤

It is obvious from the very first view when you look at it that x=5. 5 over 5 is 1. 1 multiplied by 1 is 1.

05:34 you wrote [b to_the_power log(a/b) = b] but obviously you wanted to write [b to_the_power log(a/b) = a] because this is what you substitute a bit later, correct?

X=log96/log8 X=2.194875

Or this way to get the solution (step by step) ... 7^(x + 1) + 7^(x - 1) = 12 set k = x + 1 7^k + 7^(k - 2) = 12 7^x + 7^k/7^2 = 12 7^k + 7^k/49 = 12 49·7^k/49 + 7^k/49 = 12 50·7^k/49 = 12 50·7^k = 12·49 7^k = (12·49)/50 7^k = 11.76 recall: k = x + 1 7^(x + 1) = 11.76 7·7^x = 11.76 7^x = 11.76/7 7^x = 1.68 x = ln(1.68)/ln(7) --- /// final result: ■ x = ln(1.68)/ln(7) --- /// check: 7^(ln(1.68)/ln(7) + 1) + 7^(ln(1.68)/ln(7) - 1) = 12 🙂

[7^(x+1)]-[7^(x-1)]=12 7×(7^x)+(7^x)/7=12 (48/7)(7^x)=12 (4/7)(7^x)=1 --> 4[7^(x-1)]=1 7^(x-1)=¼ Take logarithm: (x-1)log(7)=-log(4) --> x=1-⁷log(4) where ⁷log(4) is log based on 7

LHS=3^(½+¼+⅛) =3^⅞ RHS=27^x =3^(3x) Therefore 3^⅞=3^(3x) --> 3x=⅞ x=7/24 Too easy for Math Olympiad.

3^(m+9)+3^(m+14)=244 [3^(m+9)](3⁵+1)=244 244[3^(m+9)]=244 --> 3^(m+9)=1 m+9=0 m=-9

Sans passer par log 5

x = (3log343(7)) / 2 = .5

2.

2× 5^x = 2 × 5^2 Therefore, x = 2

6x + 9x = 4x + 2 6x + 9x - 4x = 2 11x = 2 x = 2 / 11 x = 0.18181

Way too many steps for a simple equation like this.

8^x + 2^x = 350 (2^3)^x + 2^x = 350 (2^x)^3 + 2^x = 350 let b = 2^x b^3 + b = 350 part proving 350 = 7^3+7 b^3 + b = 7*50 b^3 + b = 7*(49+1) b^3 + b = 7*(7^2+1) b^3 + b = 7*7^2+7 b^3 + b = 7^3+7 b = 7 2^x = 7 x = log_2(7) or log(7)/log(2)

2^x=3 2 əsasdan loqarifmləsək loq 2^x = loq3 buradan xloq2= loq3 x=loq3 8^x+2^x=(2^3)^x+2^x=2^3x+2^x= =2^3loq3 +2^loq3=2^loq3^3+2^loq3= =2^loq27+2^loq3 =27+3=30 (loqarifmlər 2 əasdandır) 2-ci üsulla həllin doğruluğunun yoxlanması: 2^x=3 olduğundan 8^x+2^x=30 tənliyində 2^x-in yerinə 3 yazsaq (2^x)^3+2^x=3^3+3=27+3=30 dəqiq olardı. Təşəkkürlər.

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Simple and nice steps to follow