@ 2:40 / 10:09 (x - 3)² - 7² = 0 (x - 3)² = 7² x - 3 = ± 7 x = 3 ± 7 → x = 10 → x = - 4 (x - 3)² + 7² = 0 (x - 3)² = - 7² x - 3 = 7².i² x = 3 ± 7i → x = 3 + 7i → x = 3 - 7i

You went way off the rails at 3:03. There is a much simpler solution than what you did. You should have factored [(x-3)^2 - 7^2] as difference of two squares: (x-3+7)(x-3-7). This would have immediately given you roots {-4,10}. Similarly you should have rewritten [(x-3)^2 + 7^2] as [(x-3)^2 - (-7^2))], also a difference of two squares, and then factored it as (x-3+7i)(x-3-7i). This would have immediately given you roots {3-7i,3+7i}.

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Great

let u=3^x , u^3+u^2+u-14=0 , (u-2)(u^2+3u+7)=0 , 3^x=2 , x=log2/log3 , test , 3^x+9^x+27^n,x=2+4+8 , --> 14 , OK , 1 -2 3 -6 7 -14

if you divide through by x^5 and substitute a=x+1/x then we get (a^2-1)/(a+1)=3 so a-1=3 a=4 x^2-4x-1=0 x=2+-ζ3

X-3=4

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There is also another result such as: x = 144 and y = 16(log6/log2)^2 You can proof check it to confirm result etc...

4^(x+2)-4^x=20 4^x(16-1)=20 4^x=4/3 x=log(4/3)/log4 x=(log4-log3)/log4 x=1-log3/2log2

x=4+3=7

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I think this ways is a bit easier: (x - 3)^6 = 4^6 (x - 3)^(3 * 2) = 4^(3 * 2) ([x - 3]^3)^2 = (4^3)^2 Let a = (x - 3)^3, and b = 4^3 a^2 = b^2 a^2 - b^2 = b^2 - b^2 a^2 - b^2 = 0 (a - b)(a + b) = 0 ([x - 3]^3 - 4^3)([x - 3]^2 + 4^3) = 0 Let a = x - 3, and b = 4 (a^3 - b3)(a^3 - b^3) = 0 (a - b)(a^2 + ab + b^2)(a + b)(a^2 - ab + b^2) = 0 (a - b)(1 * a^2 + b * a + b^2)(a + b)(1 * a^2 - b * a + * b^2) = 0 Suppose a - b = 0 a - b = 0 Remember, a = x - 3, and b = 4 (x - 3) - 4 = 0 x - 3 - 4 = 0 x - 7 = 0 x - 7 + 7 = 0 + 7 x = 7 Suppose 1 * a^2 + b * a + b^2 = 0 1 * a^2 + b * a + b^2 = 0 a = (-b +/- sqrt[b^2 - 4 * 1 * b^2]) / (2 * 1) a = (-b +/- sqrt[1 * b^2 - 4 * b^2]) / (2) a = (-b +/- sqrt[(1 - 4) * b^2]) / 2 a = (-b +/- sqrt[(-3) * b^2]) / 2 a = (-b +/- sqrt[(-1) * 3 * b^2]) / 2 a = (-b +/- sqrt[-1] * sqrt[3] * sqrt[b^2]) / 2 a = (-b +/- i * sqrt[3] * b) / 2 a = b * (-1 +/- i * sqrt[3] * 1) / 2 Remember, a = x - 3, and b = 4 x - 3 = 4 * (-1 +/- Ii* sqrt[3] * 1) / 2 x - 3 = 2 * (-1 +/- i * sqrt[3]) x - 3 = -2 +/- i* 2 * sqrt(3) x - 3 + 3 = 3 - 2 +/- i * 2 * sqrt(3) x = 1 +/- i * 2 * sqrt(3) x = 1 + i * 2 * sqrt(3), or x = 1 - i * 2 * sqrt(3) Suppose a + b = 0 a + b = 0 Remember, a = x - 3, and b = 4 (x - 3) + 4 = 0 x - 3 + 4 = 0 x + 1 = 0 x + 1 - 1 = 0 - 1 x = 1 Suppose 1 * a^2 - b * a + * b^2 = 0 1 * a^2 - b * a + * b^2 = 0 a = (-[-b] +/- sqrt[(-b)^2 - 4 * 1 * b^2]) / (2 * 1) a = (b +/- sqrt[1 * b^2 - 4 * b^2]) / (2) a = (b +/- sqrt[(1 - 4) * b^2]) / 2 a = (b +/- sqrt[(-3) * b^2]) / 2 a = (b +/- sqrt[(-1) * 3 * b^2]) / 2 a = (b +/- sqrt[-1] * sqrt[3] *sqrt[b^2]) / 2 a = (b +/- i * sqrt[3] * b) / 2 a = b * (1 +/- i * sqrt[3] * 1) / 2 Remember, a = x - 3, and b = 4 x - 3 = 4 * (1 +/- i * sqrt[3]) / 2 x - 3 = 2 * (1 +/- i * sqrt[3]) x - 3 = 2 +/- i * 2 *sqrt(3) x - 3 + 3 = 3 + 2 +/- i * 2 *sqrt(3) x = 5 +/- i * 2 *sqrt(3) x = 5 + i * 2 *sqrt(3), or x = 5 - i* 2 * sqrt(3) x1 = 7 x2 = 1 + i * 2 * sqrt(3) x3 = 1 - i * 2 * sqrt(3) x4 = 1 x5 = 5 + i * 2 *sqrt(3) x6 = 5 - i * 2 *sqrt(3)

(x - 3 )6 = 46 solution (x - 3 )6 = 46 dividing both sides of the equation by 46 ((x-3)/4)6 = 1 By taking the sixth root of the two sides of the equation (x-3)/4 = √(6&1) = 1 1/6 1 = cos 0° + i sin 0° Then (x-3)/4 = (cos 0° + i sin 0°)1/6 by applying (x-3)/4 = (cos 0° + i sin 0°)=1 Then x =7 (x-3)/4 = (cos (0+1×360)/6° + i sin (0+1×360)/6°)= 1/2 + √3/2 i Then x = 5+2 √3 i (x-3)/4 = (cos (0+2×360)/6° + i sin (0+2×360)/6°)= (-1)/2 + √3/2 i Then x = 1+2 √3 i (x-3)/4 = (cos (0+3×360)/6° + i sin (0+3×360)/6°)= -1 Then x = -1 (x-3)/4 = (cos (0+4×360)/6° + i sin (0+4×360)/6°)= (-1)/2 - √3/2 i Then x = 1 - 2 √3 i (x-3)/4 = (cos (0+5×360)/6° + i sin (0+5×360)/6°) = 1/2 - √3/2 i Then x = 5 - 2 √3 i

Substitute x=y-10 into the given equation and rearrange to 2*4(y^3+y)-544=0 or y^3+y-68=0 yields y=4 by inspection. Divide by (y-4): y^2+4y+17=0 yields y=(-4±i2√13)/2=-2±i√13. Then x=y-10=4-10=-6 or -2±i√13-10=-12±i√13

Way of explanation is very good 👌

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4

a+b=46,a^2-b^2=23=(a+b)(a-b) a-b=1/2,2b=46-1/2=91/2,b=91/4 x=(91/4)^2=8281/16

Методом подбора, за минутку😮

3^3x+3^2x=150 (3^x)^3+(3^x)^2=150 3^×=y y^3+y^2=150 y^2(y+1)=150 25×6=150 y^2=25 y=5 3^x=5 log3^x=log5 x log3=log5 x=log3(5)

4^(x+1)-4^(x-1)=5 4^2*4^(x-1)-4^(x-1)=5 4^(x-1)*(4^2-1)=5 You can exclude 4^(x+1), 4^x, 4^(x-1).... and even 4^(x+10) but then it complicates the numerical calculations. First I chose 4^(x-1). For 4^(x+10) showing below. 4^(x-1)*(16-1)=5 4^(x-1)*(15)=5 4^(x-1)=5/15=1/3 |log() both sides The essence of the solution can be applied after various transformations. log{4^(x-1)}=log{3^(-1)} (x-1)*log(4)=-log(3) x-1=-log(3)/log(4) x=1-log(3)/log(4) 4^(x+1)-4^(x-1)=5 4*4^x-(1/4)*4^x=5 4^x*(4-1/4)=5 4^x*(15/4)=5 4^x=20/15 |log() x*log(4)=log(20/15) x=log(4/3)/log(4) x=1-log(3)/log(4) 4^(x+1)-4^(x-1)=5 4^(x+10)*4^(-9)-4^(x+10)*4^(-11)=5 4^(x+10)*{4^(-9)-4^(-11)}=5 4^(x+10)=5/{4^(-9)-4^(-11)} | log() both sides (x+10)*log(4)=log[5/{4^(-9)-4^(-11)}] (x+10)=log[5/{4^(-9)-4^(-11)}]/log(4) ... In the end you will get the same answer but you have more calculations on the right side of the equation (x+10)=log[5/{4^(-11)*(4^2-1}]/log(4) (x+10)=log[5/{4^(-11)*(15}]/log(4) (x+10)=[log(5)+11*log(4)-log(15)]/log(4) (x+10)=[log(5)-log(15)+11*log(4)]/log(4) x+10=[log(1/3)+11*log(4)]/log(4) x+10=11-log(3)/log(4) x=1-log(3)/log(4)

X,y€ Z+ x+y < x^2 - xy + y^2 => WHY???? x=y=1 € Z+ , x+y > x^2 -xy +y^2 😢

3^3×+3^2×-5^3-5^2=0 =>{(3^×)^3-5^3}+ (3^×^2-5^2)=0 =>(3^×-5)(3^2×+6(5^×)+30)=0 Either ×=log@3 (5) Or {-6+_ _/6^2-120)}/2=-3+_ _/3^2-30) =(-3+__/21 i)

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3×.(3^3)^×=2(3^2) W{3×.Ln3.e^ln3(3×)} =W{2ln3.e^ln3(2)} =>3×ln3=2ln3 =>3×=2,×=2/3

(x²-3)²=x+3 →x⁴-6x²-x+6=0 →(x-1)(x³+x²-5x-6)=0 →(x-1)(x+2)(x²-x-3)=0 ∴x=1,-2, (1±√13)/2

let x=u^2 , u^2+u^2+23=46^2-96u+u^2 , u=2093/92 , x=8281/16 , test , V(8281/92+23)+V(8281/92)=93/4+91/4 , --> 184/4=46 , OK ,

Log(⁴)4/3

Your deliberation is very explicit and exact !!!!Excellent one!!!!

Lovely solution.

x^4+/-x^3-6x^2-x+6=0 , (x+2)(x^3-2x^2-2x+3)=0 , (x-1)(x^2-x-3)=0 , x=(1+/-V(1+12))/2 , x=(1+/-V13)/2 , 1 2 1 -1 solu , x= -2 , 1 , (1+V13)/2 ,(1-V13)/2 , -2 -4 -1 1 -2 -4 -3 3 3 3

Aprendi propiedades de los logaritmos que no conocia. 👏👏👏👏

Muy interesante!

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let 3^x=y , y+x=30 , x=30-y 3^x=y and x=30-y then 3^(30-y)=y [3^(30-y)=y]^1/(30-y) 3=y^1/(30-y) , 27^1/3=y^1/(30-y) 27^1/(30-27)=y^1/(30-y) result y=27 and y=3^x , 27=3^x 3^3=3^x ,. x=3

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Also x=1 is a solution

You could easily do it as follows: 4^(x+1) - 4^(x-1) = 5; or, 4^x *4 - 4^x *4^(-1) = 5; or, (4^x)(4 - 1/4) =5; or, (4^x)*(15/4). Now you can follow as you did. Dr. Ajit Thakur (USA).

x^4-2x^3-x^2-x^2+2x+1=0 x^2(x^2-2x-1)-(x^2-2x-1)=0 (x^2-2x-1)(x^2-1)=0 x^2-2x-1=0 , x=1+-√2 x^2-1=0. , x=+-1

-64x^3+64x^2-3=0 --64x^3+16x^2+48x^2-3=0 -16x^2(4x-1)+3(16x^2-1)=0 -16x^2(4x-1)+3(4x-1)(4x+1)=0 (4x-1)(-16x^2+12x+3)=0 4x-1=0 , x=1/4 16x^2-12x-3=0 x=(3+-√21)/8

2^×=4×......1/4=×.2^× =>{(-1)(1/4)(4/4) =(-4)Ln2.e^ln2(-4)} =(-×)Ln.e^ln2(-×) [On applying Lambert w-function .. =>w(ae^a)=a] =>W{-4)ln2e^ln2(-4)} =W{(-x)Ln2.e^ln2(-×)} =>-4ln2=-×.ln2 =>×=4

x^3-2x^3-2x^2+2x+1=0 , (x-1)(x^3-x^2-3x-1)=0 , (x+1)(x^2-2x-1)=0 , x=(2+/-V(4+4))/2 , x=(2+/-V(8))/2 , x=1+/-V2 , 1 -1 1 1 solu , x= 1 , -1 , 1+V2 , 1-V2 , -1 1 -2 -2 -3 3 -1 -1 -1 1

Why didn't you just sixth root it?

3^×=9× =>3^(×-2)=× =>3^-2=×{3^-×} =>-3^-2 3:08 =(-×).3^(-×) =>(-)(3)(3^-3)=(-×)(3^-×) On application of Lambert W-function W(a.e^a)=a 9:24 =>W{(-3)ln3.e^ln3(-3)} =W{(-×)ln3.e^ln3(-×)} =>-3ln3=-×.ln3 =>×=3

il risultato è x = 2 n.b. : è stata utilizzata la formula W di Lambert

3^×-1=-×+5 =>1=(-×+5)(3^-×+1) =>(3^4)ln3=(-×+5)ln3 .e^ln3(×+1+4) =>W{3ln3.e^ln3(3)} =W{(-×+5)Ln.e^ln3(-×+5)} By applying Lambert 5:57 w-function..w(a.e^a)=a =>3ln3=(-×+5)ln3 =>×=5-3=2*

3^(x-1) is monotonic increasing 5-x is monotonic decreasing. so there is only one real solution x=2 is clearly that sole real solution.

(x+3)^3=+-2^3 , (x+3)^3-+2^3=0 (x+3-+2)[(x+3)^2+-2(x+3)+4]=0 x+3-+2=0 , x=-1. or. x=-5 (x+3)^2+-2(x+3)+4=0 x+3=[-+2+-√(4-16)]/2 x+3=-+1+-i√3 , x=-4+-i√3 or x=-2+-i√3