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Can not understand a word you are saying and whats with the screaming kid. 😢

So basically just guessing 3

Too slow

Excellent. I would have used factor and remainder theorem.❤

5th root of 3 is also an answer!

x³+x = 350 x³+x = 343+7 x³+x = 7³+7 Hence, x = 7.

4ⁿ-2ⁿ=1 (2²)ⁿ - 2ⁿ = 1 (2ⁿ)² - 2ⁿ = 1 a² -a = 1; a=2ⁿ and a > 0 a²-a-1=0 a=ф 2ⁿ=ф log 2ⁿ = log(ф) n*log2= log(ф) n = [log(ф)]/[log2] or n = log2(ф) then, n ≈ 0,6942419137

A vontinuer

It's 5, I can do that in my head.

why isn't it x=1

Beautiful equation

I would have left it as log of 5 to base 2 after applying log to base 2 from the start as one log property allows canceling the log and the matching base, leaving only what is in the exponent as the power rule is implied.

Other way is: If x € Z, then: p³≈350 343<(350)<512 => 7³<350<8³ => p~>7³ r=350-7³ => t=7 --->p=r=7 =>x=7

x³+x=350; 350=2×5²×7 x(x²+1)=2×5²×7 x(x²+1)=7×50 x(x²+1)=7(49+1) x(x²+1)=7(7²+1) => x=7 1. 0. 1. -350 | 7 1. 7. 50 0 x²+7x+50=0 d½=i(151)½ x= [-7±i(151)½]/2

2 cubed is 8 so the equation is 8^x+2^x=10 where x is clearly 1

and how is that answer different from x = log (base 4) from 24 I mean I can get x = 1 + log (base 4) from 6 in one step from it but… why bother at all? One can endlessly perform change of log bases but those are just trivial operations that add no insight. This doesn’t look like an olympiad problem

Thanks

Yes your solution is right

Прикольно, молодец

Wolfram alpha says otherwise

This is a joke, right? Just by looking at it you can tell that the answer is between 1/2 and 1/3 (closer to 1/3)

very useful (im in primary school)

5^x + 25^x = 125^x 125^x - 25^x - 5^x = 0 5^x (25^x - 5^x - 1) = 0 5^x is never equal to zero for real numbers, so 25^x - 5^x - 1 must be equal zero. Let u = 5^x, u>0 u² - u - 1 = 0 D = b² - 4ac = 1 + 4 = 5 u1 = (1-√5)/2, u1<0 u2 = (1+√5)/2, u2>0 u1 doesn't satisfying our restrictions, so substituting back only u2 5^x = (1+√5)/2, therefore x = log_5((1+√5)/2)

Use substitution. Let 5^x = y Y³-y²-y=0 Y(y²-y-1)=0 y=0 or y= 1±√1²-4(1)(-1) /2(1)= 1/2 ± √5 /2 5^x=1/2+√5 /2 , 5^x=0 (rej as exponential cannot equal 0) 5^x= 1-√5 / 2 rej as exponential cannot equal negative

Well done.

4 ^x - 1 ^x =2 4 ^x = 3 = 12/4 =4 ×3/4 X = 1 ?

4^x-1^x=2^x (2^x)^2 - 2^x-1=0 Let 2^x=a So, a^2 -a -1=0 Here a=1, b=-1, c=-1 D= b^2 -4ac =(-1)^2 -(4×1×-1) =1+4=5 *D= *5 a=(-b±*D)/2a =({(-1)±*5}/2 a=(*5-1)/2 and -(*5+1)/2 The second value will produce very complex solution So a=(*5-1)/2 2^x=(*5-1)/2 Take log log 2^x=log (*5-1)/2 X. log 2= {log (*5-1) - log2 } X= {log (*5-1)-log2 }/log2 X={log (*5-1)/log2 }-1 =(log (*5-1) with base 2}-1

Nice analysis👏

Awesome bro I'm from India

x² = 9 + 4√5 x² = 4 + 4√5 + 9 (a+b)² = a² + 2ab + b² x² = (2+√5)² √x² = √(2+√5)² |x| = |2+√5| If |f(x)| = |g(x)|, then f(x) = g(x) or f(x) = -g(x) x = 2+√5, then 1/x = 1/(2+√5) Since √5 is approximately 2.24, then 2+√5 ≈ 2+2.24 = 4.24 1/x ≈ 1/4.24 ≈ 0.236 x = -(2+√5) then similarly 1/x ≈ -0.236 Answer: 1/x = ±0.236

8^x = 20 (2^x)^3 = 4 × 5 2^3x = 2^2 × 5 Divide both sides by 4 (2^3x)/(2^2) = 5 2^(3x-2) = 5 Using basic logarithmic identity 2^(3x-2) = 2^log_2(5) Bases are equal, therefore 3x - 2 = log_2(5) 3x = log_2(5) + 2 x = (log_2(5) + 2)/3 Approximately x = 1.440642698295 ≈ 1.44

56^x - 28^x = 14^x (4×14)^x - (2×14)^x - 14^x = 0 (14^x)(4^x - 2^x - 1) = 0 14^x is never equal to zero, if x∈R. Therefore let's make an equivalent transition 4^x - 2^x - 1 = 0 (2^x)^2 - 2^x - 1 = 0 Let u = 2^x, then u>0 u² - u - 1 = 0 u1 = (1-√5)/2, u2 = (1+√5)/2 Since √5 is approximately 2.2, 1-√5 will be negative. Therefore u1 does not satisfy the restrictions 2^x = (1+√5)/2 | ×2 2^(x+1) = 1+√5 Using basic logarithmic identity 2^(x+1) = 2^log_2(1+√5) Bases are equal, therefore x + 1 = log_2(1+√5) x = log_2(1+√5) - 1 Approximate answer is x = 0.694241913631... ≈ 0.7

Nice. This question seemed hard at first but only require simple theory

crazy

Just take log base 10 on either side, as log(100) = 2. x.log(4) = log(100) x.log(4) = 2 x.2.log(2) = 2 x.log(2) = 1 x = 1/log(2)

4^x = 100 <=> ln(4^x) = ln(100) <=> x*ln(4) = ln(100) <=> x = ln(100)/ln(4) <=> x = 3.321928095 If we check : 4^3.321928095 = 100 Seems simpler

nice. freaking logarithms

0:30 but are you on drugs? for real... 10 years ago since i got my last math class, i dindt even end the first course in university, and wtf are you doing with the first step??? Its that hard for you to notice that 8 = 2 * 2 * 2. this equation is super simple 2 exponent x equal to y and you have a really simple equation. y(y^2 - y -1) = 0; .... y=0 one solution and the other 2 are easy gotten from the 2 grade equation form.... what the point on adding those fractions? make it harder than it is?

I think iy would be easier if we move 5 to the side and transform it into log Like: x=log5 10 Then use calculator Is this accepted?

There's always been something strangely cathartic about watching somebody work their way through math problems in different ways. I personally would have taken a different approach. But your approach is equally valid and its really enlightening watching other people take different avenues.

stop doin that fake accent !!!!!!!!!!!!!!!!!!that's so annoyiiiing

Wouldn’t it be roughly 2.11

2ª+4ª=8ª is wrong because you can change it to (2+4)ª=8ª which is 6ª=8ª and this is wrong. (Plus you did an answer to x and not a - x is (½)ª=4, you just said that a is bigger or equal to 1 and is still incorect (i think) as 6ª is not equal to 8ª unless a is 0) (a=0 is a simple answer and a correct one) Ps.: sorry for grammar and other language problems.

Sir answer should be sqrt(2) pls check

the answer is x = 7^(1/7)

1ⁿ+7ⁿ=49ⁿ; a=7ⁿ 1+a=a² a = φ or a < 0 Solum considere a>0. Ergo: a=φ 7ⁿ=φ n*log7=log(φ) n=[log(φ)]/log7 n=log7(φ)

1ⁿ+4ⁿ=16ⁿ (2⁰)ⁿ+(2²)ⁿ=(4²)ⁿ 2⁰+2²ⁿ=[(2²)²]ⁿ 1+2²ⁿ=(2²)²ⁿ 1+2²ⁿ=(2²ⁿ)²; a = 2²ⁿ 1+ a = a² a² - a - 1 = 0 D=(-1)² - 4*(1)*(-1)=1+4=5 a=[ -(-1)±(5½)]/[2*(1)] a= [1±(5½)]/2; ϕ=(1+√5)/2 2²ⁿ=ϕ or 2²ⁿ=[(1-√5)/2] < 0 Sed, non est n| 1ⁿ= -1 et n ∈ C (But there is no n, such that 1ⁿ= -1) firmi sumus quod: 2²ⁿ > 0. Ergo: Solutio solum uti, 2²ⁿ=ϕ 2nlog2=log(ϕ) 2n=log(ϕ)/log2 2n=log2(ϕ) n=[log2(ϕ)]/2