Solving Exponential Equation

nice explanation ❤

Nice problem.

On the last step, you can reexpress ln(4) as 2ln(2) and since ln(2) ~= 0.693 is more familiar to a student and more likely memorized than ln(4), we would not need a calculator.

Thank you professor for you sharing your knowledge 👍🏻

e square power 2=e power 9

e^x -1/12e^x -1=0<=>
e^0 -12e^-0 -1=-12 <=>
e^1 -12e^1 -1=-2,69<=>
e^1,385 - 12e^-1385 -1=0,009<=>
x=1,385

I wish I had YOU as a professor!

Put directly epoer x euals a .. now its very quick solving

Excellent lesson. 👍

Is someone able to build a solution in Excel/Google? I'm having a hard time solving for a complex exponent in Google/Excel. This is based on a financial formula.

A=P*((1+r/n)^(n*t))+x

Solving this for (t)

(A-x)/P =(1+r/n)^(n*t)

At this point, I think I need to use a log function to get the exponent out, but if I capture (t) in "log(1+r/n,n*t)", I'm really not sure what to do to get (t) out of the function.

Please help!

e^x-1-12e^-x=0
(√e^x-4√e^-x)(√e^x+3√e^-x)
√e^x=4√e^-x dan √e^x=-3√e^-x
e^x=16e^-x dan e^x=9e^-x
e^x=16/e^x dan e^x=9/e^x
e^2x=16 dan e^2x=9
e^x=4 dan e^x=3

Outstanding. Succinct

Great vid, many thanks, Sir!😊

0=0

ex-12e-x-1 = 14 1 12x3=36+3=39 1.39 0 -1 ex 13x3=39 1.39-1.39=0

ex-12e-x-1=0

math

And e to the x will never be 0. If it were 0=0 , and any x is a solution... an infinite 1. But then the equation does not need be exponential.

What an excellent instructor, clear and showing all the steps and excellent shortcuts. Where were you when I was in college (40 years ago)?

I watched this 2 weeks ago. Still dont understand it fulky, but its like this cloud of uknowing slowly is become more and more clear to see through.

Man I really kind of miss studying algebra and calculus after watching your videos for a few days. Your skills are top notch. 👌 👍

Please make a video to find angle of inverse ratios(sine, cosine....)

You made it look so simple. Thank you.

Excellent! Congratulations on the clarity of your explanations!

Solving the quadratic independently, I completed the square and obtained the same result for X, discarding the negative alternative for X

Exelent

By multiplying the equation by e^x for both sides
e^2x-e^x-12=0
Let e^x=y
y^2-y-12=0
(y-4)(y+3)=0
y=4
y=-3
e^x=y=4
Take ln for both sides
xln(e)=ln(4)
x~1.4
y=-3=e^x
Rejected

Awesome, you make it look so simple. Thank you!

i wish you the best wnbk w5yrt kbire 💘💘

=(e^x-4)(e^x+3)=0=>e^x={4,}=>x=log4

Wow .. thank you sir

.e^x-12e^-x-1=0
Dikalikan dg e^x menjadi
.e^2x-e^x-12=0 yg berarti
.e^x=4 & e^x=-3
.x=log4/loge=ln4

You explain complicated problems so simply. Easy to understand!

Nice one, can you make a video about the gauss-jordan elimination

Brilliant !!

Thank you so much❤

Thank you so much❤

Nice vídeo.

If you wanted to include imaginary solutions, you could write -3 in its polar form (3e^ipi) and then take the natural log of that to get another solution, ln3+ipi

Thank you very much professor 👍

Thank You, Professor 🙌🙌🙏🙏