Can You Solve this Exponential Equation? | Easy Step-by-Step Tutorial

[iZAPMath]

You are a tremendous help to students, PreMath! Solving by substitution is hard, and I love how you explained and showed each step!

[kennethstevenson]

Thank you for reminding me that I need to practice log problems until I can use logs to solve equations. I'm 78 years old, but not to old to review knowledge I neglected in my youth. The key to learning math is practicing and utilizing skills until one is proficient. The great thing about Math is that once a skill is mastered you know that knowledge is not going to change and can be used to solve more difficult problems. Math Knowledge is a Box of Tools used to solve math problems, in order to use the tools one must practice until one knows how to use the tools.

[chiragvaghmare]

If i had youtube in my school time, i would have definately passed iit. Really great platform and great job. Keep on !

[randymartin]

Excellent thank you very much. I did not think to use u-substitution. The step by step way you showed your work and reference to rules of exponents/logs was perfect.

[r.g.]

Just so complicated! We went from a simple 4x + 6x + 9x to rocket science!!

[tombufford]

I wrote a number of computer programs with quadratic geometry equations needing to be solved. When working out the mathematics I had to remind myself of the relationships between quotients and exponentials. Well worth going over this in detail.

[tombufford]

I had a quick look at this and could not solve easily. Thank you, I found this very relaxing to follow !

[matthewjohnson]

As long as you are splitting up the multiplications in the logs to external addition, you might as well use Euler's Identity (e^pi*i = -1 : ln(-1) = pi*i) to tackle x = ln(1-sqrt(5)/2) / ln(3/2) = [Nodd*pi*i + ln(sqrt(5)-1) - ln(2)] / [ln(3) - ln(2)] = [ln(sqrt(5) - 1) - ln(2) / ln(3) - ln(2)] + [Nodd*pi/ln(3) - ln(2)]i, where Nodd is any odd integer.

You also have x = [ln(sqrt(5) + 1) - ln(2) / ln(3) - ln(2)] + [Neven*pi/ln(3) - ln(2)]i, Neven is any even integer, and your answer is for Neven = 0.

[BenZineAe]

Nice bring up more these type of Well done from heart❤️

[spacey]

In the end, when you have the log raport, I think it would look better to write it as log base 3/2 of (1+sqrt(5))/2

[JSSTyger]

For the final answer, it should be noted that you can choose ANY base for these logs BUT the base should be consistent.

[devondevon]

Answer x= 1.18681
4^x + 6^x = 9^x
4^x/6^x + 1 = 9^x/6^x ( divide both sides by 6^x)
(2/3)^x +1 = (3/2)^x
introduce y= (2/3)^x, therefore 1/y= (3/2)^x
hence y + 1 = 1/y
y^2 + y =1 ( mutliply both sides by y)
y^2 + y-1 = 0
using quadratic formulae calculator y= 0.618034 and -1.61803
Therefore (2/3)^x = 0.618034 ( I won't use -1.61803 since will log both sides, and the natural log of a negative number is undefined)
x log 2/3 = log 0.618034
-0.176091 x = -0.208987
x = 0.208987/0.17091 (since both were negative numbers)
x = 1.18681 Answer

[billybhoy]

I watched this even though I had finished university 3 years ago - just to refresh my memory :-)

[ronaldnoll]

X1= log(1.618) to base (3/2)... that's all
X1= 1.1868
A thumbs up and a thank you from me

[zyadtarek]

I just saw the video and tried solving it for an hour then i failed and saw the video till the end
You did a nice work here

[lavc]

Hello teacher, this case there is a relationship between 6 and 9, so we can use substitution, what if there is no relationship, like 7 and 9

[babysharma]

This method is more simple as compare to the method tell by mind your brain

[mastermath]

It's a question of the book, Problems in mathematics by v.govorov.

[dnarxusyt]

Indians are really good in explaining Math. Thanks!

[olivierb]

Hello,
Would you please explain me something please : You have divided by 4 ^X all the member of the equation so 4^X is the denominator but why you don't have multiply by 4^X the numerator ?? (4^X*4^X) / 4^X
Sorry for this silly question. :)