วีดีโอ

Many mistakes, and very long solution.

You say what's a? a= 1

Well explained! Thanks.

K = 2¹⁶ but why don't just tell it (2²)²

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]

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)]*

Dude, WRONG result... the correct answer is x = 0,268569. Delete this vídeo.

Makes me want to refresh my Math knowledge! Hmmm.

Bad math. In no universe does 5.2/5.5 = 2/5

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.

X = 5

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

x = 1.

I plugged the equation into my calculator and instantly got 0.23141397

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

x^3-x=24, (x)(x^2--1)=(3)(8), ×=3, ×^2--1=8, x^2=8+1, x^2=9, x=+/-3

the answer can be log(((square root of 5)+1)/2) base of log should be 7.

X=0

Nice!

pls i dont undre stand step 3

:D

Do in wrong factor.

(y)3 -(5)3 +y -5 =0; y=5

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.

Why the need for national distinction in the title?

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.

Nice to learn thank you

X=1

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. 😊

Hello. Fraction line should be in line with the "=" sign

You are very good TH-camr I wished you solve the equation but you jock to me brilliant

how come your answer is K?? how much is x???

Beautiful

your handwriting is so neat

Hocam 3.dereceden denklemler ile ilgili soru çözer misiniz?

Who would want to solve this useless ahh question 😅😂

👍

O.K

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.

Very helpful sir.

I appreciate your math skills however over 80-90% of the privileges are identical. Every problem has one number that is the square of the other leading to a simple quadratic equation😮.😮😮

2ⁿ+4ⁿ=8ⁿ 2ⁿ+(2ⁿ)²=(2ⁿ)³ Let ∆=2ⁿ ∆+∆²=∆³ ∆³-∆²-∆=0 ∆(∆²-∆-1)=0 ∆=0 2ⁿ=0 reject ∆²-∆-1=0 4∆²-4∆-4=0 4∆²-4∆+1=5 (2∆-1)²=5 |2∆-1|=√5 2∆-1=-√5 2∆=1-√5 ∆=½(1-√5) 2ⁿ=½(1-√5)<0 reject 2∆-1=√5 2∆=1+√5 ∆=½(1+√5) 2ⁿ=½(1+√5) n=log_2[½(1+√5)] ❤

Bir Türk olarak videolarınızı çok sevdim.Türkiye'den selamlar.🇹🇷

I just treated the equation like I would any other higher degree polynomial that isn’t easily factorable: set it equal to zero, use the rational root theorem, and test possible rational roots via synthetic division. After testing x=5 and discovering that it is a root, I was left with the irreducible quadratic x^2+4x+20=0. Using quadratic formula, I obtained a pair of imaginary roots: 2+4i and 2-4i. So 5, 2+4i, and 2-4i are the three complex roots of the equation. But in this case it’s pretty easy to just guess and check (if you don’t want to include imaginary solutions). So to summarize: this person’s answer is wrong.

ty

OK...don.t get this, AT ALL. I did this in my head - so, did someone mess up the video???

bro this is like a year 10 level question

If y=5^x then this simplifies to y(y^2-y-1)=0. y=0 does not give a finite value for x. Quadratic simplifies to (y-1/2)^2-(sqrt(5)/2)^2 Solution y= 0.5×(1 +/- sqrt(5)) , i.e. 1.618034 and -0.618034 And x=ln(y)/ln(5), hence the only solution for y that gives a valid solution for x is y=0.5(1+sqrt(5)) =1.618034 ... of course the golden mean, and hence x=0.29899 And calculator check 125^0.29899-25^0.29899-5^0.29899=0 checks out.

In fact, it can be solved with 6 steps.

Why? There is an easier way, with less than half the steps.