x=3

3

Why do you split log10. Log10 is 1.

beautiful work!

I've solved Golden Ratio problems like this so many times I can do it in my sleep. a=ln[(√5+1)/2]/ln2=0.694241913630617...

“One” is pronounced “wun” /wan/. Rhymes with “fun”, “run”, “sun”, and “done”.

thanks

Why not simply let 2^a=k from the start ?just put 1^a = 1 (for all a) and 4^a = (2^a)^2 and you get the same equation in k^2=1+k but in the second row

No need to divide because 324=18^2 so original equation is 1+18^x=(18^2)^x so 1+18^x=(18^x)^2 . Let 18^x=k so 1+k=k^2 so k^2-k-1=0 so D=(-1)^2-4•1•(-1)= 1+4=5, So k=(+1+√5)/2 or k=(+1-√5)/2 so 18^x=(1+√5)/2>0 Accepted or 18^x=(1-√5)/2<0 Rejected. Etc .. taking Log Both Sides.

x^3-x=24 Very simple. 3^3-3=24.

Factor out x to x(X^2 - 1) = 24 where X^2 - 1 factors to (X-1)(X+1) then you have (X-1)*x*(X+1) = 24. This means the product three consecutive numbers is 24. Brute force it to x = 3 where x-1 = 2 and x+1=4 and 2*3*4 = 24

I really don't understand these methods that require remembering so many formulae. In this case, the factors of a "difference of cubes." By the time you've observed the relationship between 3, 24, and 27, you've already found a real root, specifically 3, since 3^3 - 3 = 24. Then, just long division of x - 3 into x^3 - x - 24 and use the quadratic formula, to find the complex roots.

3

let u=2^x , 738=2*3*3*41 , u^3+/-u^2+u-738=0 , (u-9)(u^2+9u+82)=0 , 2^x=9 , x=log9/log2 , test , 8^x+2^x=729+9 , --> 738 , OK , 1 -9 9 -81 82 -738

very nice ... thank you

Easy,1a+7a=49 so 49/8=6.125 ,a=6.125

k=(log(sqrt5+1)-log2)/2log2=log(sqrt5+1)/2log2-½ k≈0.3471 4^0.3471+16^0.3471≈4.236 64^0.3471≈4.236

The internet has hundreds of Golden Ratio problems like this one, and I can now solve them in my sleep. For example, a=ln[(1+√5)/2]/ln2 without writing anything down. OK, I recognized that this problem may be converted into the Golden Ratio form Φ^2-Φ-1=0, which has positive root Φ=(1+√5)/2. To obtain this form divide the given equation by 7^a and rearrange to (2^a)^2-2^a-1=0. Then substitute Φ=2^a to obtain the Golden Ratio equation and finish up, a=lnΦ/ln2.

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10x = 130 x = Log 130 / log10 x = 2.11394 10^2.11394 = 130

10k + 100k = 1000k 110k = 1000 k = log 1000 / log 110 K = 1.46958 110^1.46958 = 1000

It is just log6(50)

X=log6(50)=log6(5²*2)=log6(5²)+log6(2)=2log6(5)+log6(2)

You really don't need this hard of a solution. log² (2^3x) + log ² (2^x) = log² 30 3x + x = log² 30 4x = log² 30 x = (log² 30) / 4

Log(24)/log(8) ?

Idk how I've found your channel but i'm an engineer and I just love clasic math equations from my highschool days done step by step as a tutorial, gave me a bit of nostalgia, subscribed

8ˣ + 2ˣ = 68; 2³ˣ + 2ˣ = 64 + 4 = 2³*² + 2²; 2³ˣ = 2³*²; 2ˣ = 2²; x = 2;

It's actually 1, uh...

nice solution

You could just make everything same base and remove them. Solving the exponents as a simple ecuation

Curious. Applying the Log with base 2, e.g. Log2(2) = 1, immediately gives 3m.Log2(2) = Log2(20) and m = Log2(20)/3 = 1.4406. It’s not worth watching all these disgusting promotion videos, which even take more time than your lengthy solution.

1^x+7^x=49^x 1+7^x=(7^x)^2 K=7^x 1+K=K^2 K^2-K-1=0 . . .

Why was it necessary to make 2^x=K? 2^x . 2^x . 2^x= 2^3x. if 2^3x = 2^4 which is 16, then 3x=4 and x=4/3.

The question at the beginning was k=? Now, what is the value of k? It took me about two minutes and five lines of writing (including the original line and the final result) to get k=0.694… So what is all this idiotic calculation for?

100

5 just calculate a few small nos

You can't just reject the negative solution to the 4th power unless the question states "find the real solutions" or similar. There are at least k complex solutions to the equation.

Veryyyy good explanation...keep it Up!!! Love It!!!<3

Won't it be -2+4i for the second value of x?

X^3 -X^2 = 100 X^3-2=100 X =100

Very good explanation

why dont you use log in base 2 right away when you switch to logarithms? In other words convert 2^3x=18 directly into 3x=log[in base2]18 because the log in base 2 of 2^3x is obviously the exponent, i.e: 3x. You would reach the same solution faster.and more intuitively. Is there any rule that forces you to use log in base 10 at first?

Nice job. I only mean to help your channel succeed when I tell you (a) watching you re-over-draw what you already wrote perfectly is why I will actually avoid your videos. (b). "x = l2(3)" where I write l2() to indicate "log base 2" gets points for compactness, but falls short of *elegance*. Remember, our calculators do not usually have a button for that. IMHO answers must be presented in "calculator friendly fashion", which is "3 ln / 2 ln" in this case, or in typeset x = (ln3)/(ln2). All the other equivalent suggestions, no matter that yours are very clever, hinder practicality.

Nice ❤ Keep it up Support from india ❤

Unnecessarily long...log100 is simply 2 so it simplifies as 1+ 2/0.9 =3.2222