Too many steps. You just need to substitute a=3^x, then You get a^2-a-1=0. Then you get 3^x= (1+/-2.23)/2. Then x is either log1. 61/log 3 or log - 0.61/log 3.
What about this? 2ª + 4ª = 8ª .......... *(a = ?)* 2ª + (2²)ª = (2³)ª (2ª) + (2ª)² = (2ª)³ *Let: (x = 2ª)* x³ - x² - x = 0 (x) (x² - x - 1) = 0 D = b² - 4ac = (-1)² - 4(1)(-1) = 1 + 4 = 5 (- b) ± √(D) 1 ± (√5) 1 x = -------------------- = -------------- = ----- (1 ± (√5)) (2)(a) 2 2 *Well, we have 3 value of x* x₁ = 0 x₂ = ½ + (½)(√5) x₃ = ½ - (½)(√5) *Let's find the value of a* x₁ = 2ª 0 = 2ª a = ²log(0) = undefined, *so ignore it, it's not the answer* x₂ = 2ª ½ + (½)(√5) = 2ª a = ²log [(1+√5)/(2)] a = ²log(1+√5) - ²log(2) *a = ²log(1+√5) - 1* x₃ = 2ª ½ - (½)(√5) = 2ª a = ²log [(1-√5)/(2)] a = ²log(1-√5) - ²log(2) a = ²log(1-√5) - 1 *but (1-√5) is equals to negative, while negative number doesn't allow under log, so ignore this too, it's not the answer* *So, the answer is* *a = ²log(1+√5) - 1*
@espadadedosfilosYHWH17 күн бұрын
You say what's a? a= 1
@halal_baconhair18 күн бұрын
Well explained! Thanks.
@Nguyễn-j9q19 күн бұрын
K = 2¹⁶ but why don't just tell it (2²)²
@kksomdr19 күн бұрын
1^x + 9^x = 81^x. Let 9^x = y, y>0, 1^x = 1. 1+ y = y^2. y^2 - y - 1 = 0. y = (1 + root5)/2. y = 9^x = (1 + root5)/2. Therefore, x = log [base : 9 , exp (1+root5)/4]
@TidakTerdefinisi19 күн бұрын
I use another way to solve this problem. 1ª + 8ª = 64ª --------> *Question: a = ?* 1ª + (2³)ª = (2⁶)ª 1ª + (2ª)³ = (2ª)⁶ 1ª + (2ª)³ = [(2ª)³]² * Let: y = (2ª)³ *1ª + y = y²* Now, how to substitute (1ª) ? How about this method? y = (2ª)³ = (2³)ª For aⁿ ÷ bⁿ = (a/b)ⁿ, so we know that (2³)ª ÷ (2³)ª should be equals to [(2³)/(2³)]ª = (1)ª, therefore we can substitute (1ª) with [(y)/(y)]. And as we know (y÷y) should be equals to 1, well we can simply say that (1ª) = 1. Now, we can write the equation above as 1 + y = y² y² - y - 1 = 0 D = b² - 4ac = (-1)² - 4(1)(-1) = 1 + 4 = 5 y = [(-b) ± √(D)] / (2a) = [-(-1) ± √(5)] / (2) = ½ ± (½)(√5) Because y = (2ª)³, so (½) [1 ± (√5)] = (2ª)³ (2⁻¹) [1 ± (√5)] = (2)³ª [1 ± (√5)] = (2)³ª⁺¹ Thus, 3a + 1 = ²log(1 ± (√5)) 3a = ²log(1 ± (√5)) - 1 3a = ²log(1 ± (√5)) - ²log(2) 3a = ²log([1 ± (√5)] / 2) a = (⅓) [²log([1 ± (√5)] / 2)] Because log of negative number is undefined, then *a = (⅓) [²log([1+(√5)]/2)]*
@jorgerio688821 күн бұрын
Dude, WRONG result... the correct answer is x = 0,268569. Delete this vídeo.
@orchidorio21 күн бұрын
Makes me want to refresh my Math knowledge! Hmmm.
@shadowlord722 күн бұрын
Bad math. In no universe does 5.2/5.5 = 2/5
@joeldoxtator980424 күн бұрын
This proof seems to be missing a step. It is unclear how you went from a=log1/2((5^1/2 -1)/2) to 8^5 + 8^5 = 2 or why.
@eftekhar-naeem24 күн бұрын
X = 5
@sushileaderyt195724 күн бұрын
Just graph it out? Come on, it’s 2024 and we have a lot of compute power, stop using your brain immediately and use calculators for solutions
@Nikioko24 күн бұрын
x = 1.
@dhy534226 күн бұрын
I plugged the equation into my calculator and instantly got 0.23141397
@epsi26 күн бұрын
10:58 a = log_2((1 + √5)/2) Solution with a fraction base is not incorrect, but you can write as integer since it is 1/b = b^(-1). Use reciprocal of log base and log input and remove sqrt from denominator of log input. If you don't believe: a•log(1/b) = log(c/d) a(log(1) - log(b)) = log(c/d) a(0 - log(b)) = log(c/d) -a•log(b) = log(c/d) a•log(b) = -log(c/d) a•log(b) = log( (c/d)^(-1) ) a•log(b) = log(d/c) a = log_{b}(d/c) b = 2 , c = (√5 - 1) , d = 2 in this problem
the answer can be log(((square root of 5)+1)/2) base of log should be 7.
@TheSamuel1365Ай бұрын
for this you need to solve the equation a^2-a-1=0
@Nguyễn-j9qАй бұрын
@@TheSamuel1365 ok i forget that
@Nguyễn-j9qАй бұрын
X=0
@TheSamuel1365Ай бұрын
1+1=2?????
@fullc0deАй бұрын
Nice!
@user-vi5jd8mw5dАй бұрын
pls i dont undre stand step 3
@Nguyễn-j9qАй бұрын
:D
@Yash-ir9kuАй бұрын
Do in wrong factor.
@wieskrz8490Ай бұрын
(y)3 -(5)3 +y -5 =0; y=5
@tungstentoasterАй бұрын
Neat! You could also expedite this solution by using (8^0)^x + (8^1)^x = (8^2)^x as your first substitution. No fractions required.
@HeckaSАй бұрын
Why the need for national distinction in the title?
@MrGeorge1896Ай бұрын
But why so complicated? We can start with y = 7^x as the first step and get 1 + y = y² and so y = (1 + √5) / 2 with x = [ ln(1 + √5)- ln 2 ] / ln 7.
@saraswatinagarajan5315Ай бұрын
Nice to learn thank you
@k.kranasiri1882Ай бұрын
X=1
@alvarofernandez5118Ай бұрын
I suggest substitution of 6^x for y, rearrange and you get to the quadratic equation you need faster. Then I prefer to take the log base 2 of both sides when you solve for x, and turn the fraction into a difference of logs. One of these is log base 2 of 2, so it simplifies to 1, and the final answer is log_2 (sqrt(5) -1) -1, if my math checks out. 😊
@gabrielc6252Ай бұрын
Hello. Fraction line should be in line with the "=" sign
@kennybayudan5616Ай бұрын
You can hardly align written work.
@arifadem9745Ай бұрын
You are very good KZbinr I wished you solve the equation but you jock to me brilliant
@crieliocrielАй бұрын
how come your answer is K?? how much is x???
@nightwhenjarАй бұрын
Beautiful
@mystic-cherry-manАй бұрын
your handwriting is so neat
@okkescinar1777Ай бұрын
Hocam 3.dereceden denklemler ile ilgili soru çözer misiniz?
@focusedfox7167Ай бұрын
Who would want to solve this useless ahh question 😅😂
@PataravitPrapapuwamasАй бұрын
👍
@jagadiswarchakraborty295Ай бұрын
O.K
@tombufford136Ай бұрын
At a quick glance I start this by writing the equation 2^a + 2^2a = 2 ^3a. I see this already in the comments below.