วีดีโอ

Not fully correct, forgot the + 1 at the end.

Tanks for watching

the solution is x=log_7 (1+√5)/2

x=log((1+√5)/2)/log(7) is the only real solution: 49^x-7^x-1^x=0 is equivalent to (7^x)^2-7^x-1=0. This is a quadratic equation in 7^x. Use the formula for quadratic equations to find 7^x=(1±√5)/2. From this we have x=log((1±√5)/2)/log(7), but only the positive sign gives a real solution.

At 4:45 why do you put the 1 inside the bracket? Shouldn't it be on the outside?

Seven. Took me about 10 seconds in my head. Since we are dealing with a cube, it has to be a single-digit number. For example, 10 cubed is 1000, way too large. Five cubed is 125, six is 216, seven is 343. Aha! Now add 7 to that and you have 350.

A little bit too difficult for me. But thanks for the interesting problem and your clever solution.

beautiful content

CAN U PLEASE EXPLAIN WHEN DO WE USE LOG

On the valid solution after getting into the b^x=y form, it is best to use the log base that matches the base for the exponential. That way, you can solve the exponential equation with a single step canceling the log with the matching base as the exponential, leaving only whatever is in the exponential part. The power rule for logs is implicit whenever you use the appropriate base when applying logs at the start.

In the first one you can represent it like:: (x-1)x(x+1) = 24 which means basically find the three numbers that come after one another that their multiplication gives 24, which is a little easier to see at least for me

The value of x is 3 bro it took 3 seconds to figure out

5^x=250 => x=log5(250). Easy.

Asik asik...

x =5

Tank your video

Divide by 2^a. Then equation is x^2-x-1=0, where 2^a=x. Hence 2^(1+a)=1+sqrt(5). Ans. a=0.696 (approx.)

If a=0, then LHS=2 and RHS=1, thus LHS>RHS. If a=1, then LHS=6 and RHS=8, thus LHS<RHS. Since we need LHS=RHS, thus 0<a<1. If a=1/2, then LHS=3.41 and RHS=2.83, thus 0.5<a<1.

Correct answer should be x = log↓2(3) Or x = log(3)/log(2)

100 = 10^2. 100^x = 10^2x. 1^x = 1; The equation becomes 1 +10^x = 10^2× Let y =10^x the result is 1 + y = y^2 or y^2 -y-1 = 0, a quadratic equation. The positive root is y = 1.613 approx. X = log 1.6 or 0.20763 Ans

Very good

Good one

Intresting problem. Keep going man!

That's so satisfying

Nice video buddy... Keep it up. I enjoyed watching this

8^x is simply 2^x cubed. The two numbers have to add to 130. Therefore, these numbers must be 125 and 5. So 2^x = 5 -> x = log₂5

Answer is 3 .X=3.

There is no need to calculate: 2

Xis =1.

but if 5^3,43070 it equals =250 which x4 = 1000 right?

I've put the solution, x=6/5 or x=1.2 back into the original eqn and got 1 + 8.5858 = 73.7162 which is obviously incorrect. What have I done wrong?

Divide all the terms by 10^x, and you will have 1 = y^2 - y, where y=10^x. It is a neater way to find the answer. Btw, the answer is x= log10[(1+√5)/2]

You are wrong.

y=6 => x^3=6 => x=6^(1/3)

That's A LOT of work! Much simpler to rewrite the equation to x=log (base 3) of (argument 30) or approximately 3.0959. You're going to need a decent scientific calculator for that of course but since you're going to need a calculator ANYWAY, you may as well get a decent one. In fact I found a free app that I placed on my phone which even handles imaginary numbers! With that being said being able to understand the logarithmic rules is important. Also, if I may be so bold, get a manicure before you post another video. Handwriting was legible but taking care to form all numbers and symbols accurately without retracing them would make a much better video. This I believe is especially important for POCs because there just aren't enough of them making videos about math and with TH-cam being as crowded as it is a person really needs to pay attention to small details while providing great content.

من عدد هفت چهارم رو بدست اوردم یا لگاریتم گیری از طرفین و فاکتور گیری...❤ که تقریبا با فکر کنم ۳ صدم خطا با جواب تو برابری میکنه!😂 چون لگاریتم هست دیگه عدد ها که کاملا صحیح نیستن

log5(10)

writing your x as a backwards n should lead to a prison sentence.

Really nice

Great solution!

a=ln(golden ratio)/2

1/2 by inspection. 4^(1/2) = 1/(1/2) --> 2 = 2

Thank youuu

For the first question, it would be easier to just say x = log(40)/log(5). You went wrong when you jumped from 1+log8/log5; the fraction component is not easily to simplify and certainly cannot be simplified by changing the base of the logs. Idk why everyone else is being so mean in the comments section, it was just a mistake. You can avoid these in future by checking that your answer for X satisfies your original equation. In this case, take your answer x = 1+log5(8), and check if 5^x gives 40. Clearly it doesn't, so you know you've gone wrong somewhere.

1 question worng 2nd one i think is right

bro whatever maths ur teaching i humbly request u to stop it ur teaching maths wrong even if ur teaching something teach it correctly the method is wrong

The latter part of the video is even wierder. What the heck are you doing? This is simple mathematics and you are doing as if everybody is nuts. 😮

A simple and humble log will do. Stop trying to be smart, will you?

You had a good idea to transform ln(6*3) to ln6 + ln3. But when you divided by ln6 the rest of the algebra was incorrect, you did multiple errors.

bruh ur brain functions so good keep it up 👌🏻👌🏻