It always amuses me when you say “let's do the second method first“ - though it might be the second method you thought of. Can it be anything but the first method presented?
I did the first method (2nd shown in the video). Enjoyed the video; I mean, it's pretty simple but does keep my brain in math mode (I'm in 12th grade and I'm in the autumn break, but I enjoy maths nevertheless)
I used log(x) instead of ln(x). The answer in the video was (simplifying) ln(3)/ln(5) -3 , but this must be the same as log(3)/log(5) -3. This implies ln(3)/ln(5) = log(3)/log(5). Using the the change of base rule to convert to natural log we get log(3)/log(5) = ln(3)*ln(10)/ln(5)*ln(10) = ln(3)/ln(5). The base you choose is irrelevant.
Take the log on both sides of the equation, which is xlog(3)=(x^2+3x)log(5). Using distributive property and setting up as a quadratic equation turns out to be x^2log(5)+3xlog(5)-xlog(3)=0. Factor out x, which is x(xlog(5)+3log(5)-log(3))=0. This means that x=0 and xlog(5)=log(3)-log(125)=log(3/125), or x=log_5(3/125).
I love the content you post and I watch and learn a lot from your channel. Keep It Up! Just 1 question:- we found that x=0. But, in the equation 3=5^x+3, if we substitute x with 0, we get 3=5^3, 3=125. How come?? Edit:- Just realized, how did you multiply the exponents of both sides by 1/x as x=0 and 1/0 is meaningless??
The constant 3 raised to x will never be equal to a higher constant raised the same power squared and added to3 times X Regardless of methods solved This conclusion based on the way it is written and application
I overthought and used completing the squares, when I could just factor out x and separate the cases. Well, I got the answer correct anyway, just the very time-consuming way of finding it. 🥲
It always amuses me when you say “let's do the second method first“ - though it might be the second method you thought of. Can it be anything but the first method presented?
I think that nobody ever had that stupid jokes
@@nikolakosanovic9931
One of my mathematics teacher in secondary school was very fond of saying “watch the board while I run through it“
I did the first method (2nd shown in the video).
Enjoyed the video; I mean, it's pretty simple but does keep my brain in math mode (I'm in 12th grade and I'm in the autumn break, but I enjoy maths nevertheless)
Too easy Syber, I already knew this was going to have two solutions.
You are 🔥
Love it ...when you start with the second method
Glad you liked it!
@8:00 It's a bucket
Given:
3↑x = 5↑(x² + 3x)
To find:
x
taking natural logarithm of both sides:
ln(3↑x) = ln(5↑(x² + 3x))
Using ln(a↑b) = bln(a):
xln(3) = (x² + 3x)ln(5)
Bringing all terms to LHS:
x²ln(5) + 3xln(5) - xln(3) = 0
x(xln(5) + 3ln(5) - ln(3)) = 0
From here, either x = 0, or xln(5) + 3ln(5) - ln(3) = 0
xln(5) + 3ln(5) - ln(3) = 0
xln(5) = ln(3) - 3ln(5)
x = ln(3)/ln(5) - 3.
x = 0.
I used log(x) instead of ln(x). The answer in the video was (simplifying) ln(3)/ln(5) -3 , but this must be the same as log(3)/log(5) -3. This implies ln(3)/ln(5) = log(3)/log(5).
Using the the change of base rule to convert to natural log we get log(3)/log(5) = ln(3)*ln(10)/ln(5)*ln(10) = ln(3)/ln(5). The base you choose is irrelevant.
I would also like to point out that ln(3)/ln(5)=log5(3)=log(3)/log(5)
base 10 vs base e, just use base 5 in this case. x = log5(3) - 3
Got 'em both!
Great video! thank you sir
Why use ln() while you can use log_5(3)?
You used log of base 5 on both sides?
Take the log on both sides of the equation, which is xlog(3)=(x^2+3x)log(5). Using distributive property and setting up as a quadratic equation turns out to be x^2log(5)+3xlog(5)-xlog(3)=0. Factor out x, which is x(xlog(5)+3log(5)-log(3))=0. This means that x=0 and xlog(5)=log(3)-log(125)=log(3/125), or x=log_5(3/125).
Sir, if we take log to base 5 on both sides, we can get the answers easily.
Method 3: raise each side to the power of 1/x, then take the ln of both sides.
We can take log also instead of Ln.. or log to the base 3 or 5
Just at the starting tale the log base 5...
Can you solv
Log10(x^2 _ 3)= log5(8x) log
Note : You can have any base instead of ln
I love the content you post and I watch and learn a lot from your channel. Keep It Up!
Just 1 question:- we found that x=0. But, in the equation 3=5^x+3, if we substitute x with 0, we get 3=5^3, 3=125. How come??
Edit:- Just realized, how did you multiply the exponents of both sides by 1/x as x=0 and 1/0 is meaningless??
We can only do it if x does not equal zero
The curve 5^(x^2+3x) is symmetrical wrt x=-3/2, as it would be x^2+3x.
Nice graph. Looks like a bottle bottom.👍
can you solve this
Q. log₄[8^(3 + 1/3log₂x)]^1/3 = 4 then x = ?
Solution:- th-cam.com/video/f0_48zzmqEM/w-d-xo.html
x=0, -3+log_5(3)
The constant 3 raised to x will never be equal to a higher constant raised the same power squared and added to3 times X
Regardless of methods solved
This conclusion based on the way it is written and application
Solving an Interesting Exoponential Equation: 3^x = 5^(x^2 + 3x); x = ?
Convert the exponential base number 3 into 5 using the logarithmic math:
Let: 5^n = 3, n = log3/log5 = 0.683; 5^0.683 = 3
3^x = 5^0.683x = 5^(x^2 + 3x); x^2 + 3x = 0.683x, x(x + 2.317) = 0
x = 0 or x + 2.317 = 0; x = - 2.317
Answer check:
x = 0, 3^x = 3^0 = 1 = 5^(x^2 + 3x) = 5^0; Confirmed
x = - 2.317, 3^x = 5^0.683x = 5^(x^2 + 3x); Confirmed
Final answer:
x = 0 or x = - 2.317
We want a hard exercices pleez🙂🙂
can you solve this
Q. log₄[8^(3 + 1/3log₂x)]^1/3 = 4 then x = ?
Solution:- th-cam.com/video/f0_48zzmqEM/w-d-xo.html
log(5)3-3=-2,3174...oltre la soluzione banale x=0
this is the simplest way
5^(log5(3)x) = 5^(x²+3x) -> log5(3)x = x²+3x -> x = log5(3) - 3, x=0
Nice
I love both ways!!!
Glad to hear that!
No necessity to take natural logarithm.
x = 0. 1 second to solve.
x = 0
X=0.
ln!
I overthought and used completing the squares, when I could just factor out x and separate the cases. Well, I got the answer correct anyway, just the very time-consuming way of finding it. 🥲