3^x + 6^x = 12^x so 3^x + 2^x 3^x = 4^x 3^x so 1 + 2^x = 4^x so 4^x - 2^x -1 = 0 so 2^x = [1+-sqrt[1+4]]/2 so x = log[[1+-sqrt5]/2]/log2

Nice

Estou assistindo do Brasil 🇧🇷🇧🇷🇧🇷

I am actually surprised that Harvard has such easy questions on their entrance test

This is why I can't excel in maths :)

nice vid, just try to make the vid ratio 16:9, not 4:3

You complicated this wayyyy too much

I am too dumb now ( its only been 10 months for me to forget everything)

You complicated this equation. 1^x=1, so (8^x)²-(8^x)-1=0 Now solve for 8^x (which can't be negative) and then take logarithm on both sides to obtain the value of x.

Sorry my ignorance, but why log (root5 -1)/2 over log 1/8 is equal to log (root5 -1)/2 in base of 1/8? I mean, I was only familiar to the law that you could subtract these 2 logarithms Like log (root5 -1)/2 - log (1/8)

great video 💗

= log3(8)

X,^3-x^2=x^2(x-1) X^2(x-1)=100=25×4 =5^2(5-1) x=5

Bro please explain this rule how is apower a = n power n

Thank You Sir from India

To be honest you should solve quickly and more than 2 problem in one video This is just my opinion 😅

8^4+2^4=30????

the actual answer is 1.58 btw..

I think it's a bit easier to rewrite the equation as: (10⁰)ˣ + (10¹)ˣ = (10²)ˣ then (10ˣ)⁰ + (10ˣ)¹ = (10ˣ)² then (10ˣ)² - (10ˣ)¹ -1 = 0 Now, you have a quadratic in disguise with 10ˣ, which you solve using the quadratic formula. This gets you 10ˣ = (1+√5)/2 as you only take the positive solution. Log both sides and get your answer without the awkward base 1/10 log. Nice video though, your solution is very creative.

Solution: 49ˣ = (7²)ˣ = (7ˣ)² 42ˣ = (6 * 7)ˣ = 6ˣ * 7ˣ 36 = (6²)ˣ = (6ˣ)² divide both sides of the equation by 7²ˣ and you get: 1 - (6/7)ˣ = (6/7)²ˣ 1 - (6/7)ˣ = ((6/7)ˣ)² substitute (6/7)ˣ = t 1 - t = t² |+t - 1 t² + t - 1 = 0 t = -1/2 ± √ ((1/2)² - (-1)) t = -1/2 ± √ (1/4 + 4/4) t = -1/2 ± √5/2 t = (-1 ± √5)/2 since (6/7)ˣ can never be negative, t has to be positive, as such: t = (-1 + √5)/2 resubstitute: (6/7)ˣ = t (6/7)ˣ = (-1 + √5)/2 |log x * log(6/7) = log((-1 + √5)/2) |:log(6/7) x = log((-1 + √5)/2) / log(6/7) x = (log(√5 - 1) - log2) / (log6 - log7) x = (log(√5 - 1) - log2) / (log2 + log3 - log7) x ≅ 3.1216977...

Why didn't you change log(1/10) into -1? log(1/10) = log(10^-1) = -1*log(10) = -1 as log(x) is simply 10-based logarithm not the natural log

nice

I dont understand the rule in 1:36. why is if a powered a = n powered n same as a=n? Can you show me an example with real numbers?

Nice videos brother !!

Thanks for the explanation.🥰👍👍

nice

nice

Respected Sir, Good evening

^= read as to the power *= read as square root As per question 2^a+4^a=8^a 2^a+(2^a)^2=(2^a)^3 Let 2^a=t So, t+t^2=t^3 t(1+t)=t^3 1+t=t^3/t 1+t=t^2 t^2-t-1=0 Here, a=1, b=-1, c=-1 D=*{(-1)^2-4.1.(-1)} =*{1+4)=*5 t=(-1±*5)/2 2^a =(-1±*5)/2 Take log log 2^a=log (-1±*5)/2 a. log2 ={log (-1±*5)}-log 2 a=[{log (-1±*5)}-log 2]/log2 (may be )

🦦💪👍🩷

Set y = 2^x y^3+y= 30 y* (y^2+1)=30 y = 3 3* (9+1)= 30 3=2^x Log 3 = x Log 2 x =Log3/Log2

Странно Если 8=2^3, 4=2^2 и замена 2^а=х то х+х^2=х^3 х^2-х-1=0 И все, дальше просто Зачем какие-то дроби делать

x should be equal to 1 2³x +2x 2³+2¹ = 8+2=10

the answer is wrong

Handwriting so neat...

Also known as: log (base 6) of phi, where phi is the golden ratio.

Why overcomplicate the first one so much? Just substitute for y=5^x and solve accordingly. You avoid negative logs and the whole first division part which doesn't really achieve anything.

Thanks for this. I love your clear reasoning and explanation. Wonder if an improvement would be to substitute u = 6^x? This avoids some of the fractions. Then, as 36^x = u^2x, you can write the quadratic as u^2 - u - 1 = 0. Solving this with quadratic formula gives u = { 1+ sqrt(5) } / 2 , plus a negative answer which can be rejected. 6^x = { 1+ sqrt(5) } / 2 log 6^x = log [ { 1+ sqrt(5) } / 2 ] x log 6 = log [ { 1+ sqrt(5) } / 2 ] x = log [ { 1+ sqrt(5) } / 2 ] / log 6 x = log6 [ { 1+ sqrt(5) } / 2 ] Also... { 1+ sqrt(5) } / 2 is the golden ratio... a nice little Easter egg they slipped in there.

Can You Press Harder On The Paper Till Your Pen Breaks?

Divide by 9^X in first step would've made it simpler😅

You can use y=6^x to save some time.

thanks khubaeb

1^x=1

Recognize this as a Golden Ratio problem for easy solution. Divide the given equation by 4^x and rearrange to (4^x)^2-4^x-1=0. Substitute Φ=4^x to obtain Φ^2-Φ-1=0, which has positive root Φ=(√5+1)/2 and solution x=lnΦ/ln4.

100^100=(10^2)^100=10^200=10^(100+100). Thus, x=100.

X=log3/log2

I got x = (ln((1 + √5)/2))/ln(2) ≈ 0.694241 and your answer is ⁵√10 which is 1.584893 (not matching with 36ˣ - 9ˣ = 18ˣ). This way: 36ˣ - 9ˣ = 18ˣ (36ˣ - 9ˣ)/9ˣ = 18ˣ/9ˣ 4ˣ - 1 = 2ˣ 4ˣ - 2ˣ - 1 = 0 (2ˣ)² - 2ˣ - 1 = 0 set k = 2ˣ k² - k - 1 = 0 /// quadratic formula: Δ = (-1)² - 4·1·(-1) = 1 + 4 = 5 √Δ = ±√5 • k = (-(-1) + √5)/(2·1) = (1 + √5)/2 • k = (-(-1) - √5)/(2·1) = (1 - √5)/2 // root #1: 2ˣ = k = (1 + √5)/2 => 2ˣ = (1 + √5)/2 2ˣ = (1 + √5)/2 ln(2ˣ) = ln((1 + √5)/2) x·ln(2) = ln((1 + √5)/2) x = (ln((1 + √5)/2))/ln(2) / check: • 36^((ln((1 + √5)/2))/ln(2)) - 9^((ln((1 + √5)/2))/ln(2)) = 7.438113 • 18^((ln((1 + √5)/2))/ln(2)) = 7.438113 // root #2: 2ˣ = k = (1 - √5)/2 => 2ˣ = (1 - √5)/2 2ˣ = (1 - √5)/2 ln(2ˣ) = ln((1 - √5)/2) note: • ln((1 - √5)/2) is a logarithm of a negative number • => root #2 is a rejected solution /// final result (one root): ■ x = (ln((1 + √5)/2))/ln(2) ■ x ≈ 0.694241 🙂

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The way you solve this problem for a is so long.

Dreadful presentation. Bad technique, bad English, and deceptive title. Anyone with basic math can see the real root instantly and mentally. This presenter is so sloppy he can't even be bothered to use fresh sheets of paper. YUCK! Beside the trivial real solution there are 8 complex solutions not mentioned. Among them are solutions that 8% are unlikely to get right, for example: x = -(3/2)^(2/3) sin(1/3 (tan^(-1)(sqrt(2)) - π)) - (3^(1/6) cos(1/3 (tan^(-1)(sqrt(2)) - π)))/2^(2/3) + i ((3/2)^(2/3) cos(1/3 (tan^(-1)(sqrt(2)) - π)) - (3^(1/6) sin(1/3 (tan^(-1)(sqrt(2)) - π)))/2^(2/3))