Squaring both sides of the given equation, then dividing thru by 4^x, term-by-term, we get: (3/4)^x - 1 = [(3/4)^x]² + 1 - 2(3/4)^x. Letting u = (3/4)^x, and collecting all like terms to the right side of the equation, we get: u² -3u + 2 = 0 (u-1)(u-2) = 0 so u = 1 or u = 2. Back-sub: (3/4)^x = 1 => x = 0 and (3/4)^x = 2 => x = ln2/[ln3 - ln4].
I got 0 and log base 4/3 of 1/2... set a=3^x and b=4^x (and not b=2^2x which is a complication as 4 is a factor of both 12 and 16). substitute and square both sides to get: ab-b^2=a^2-2ab+b^2, which simplifies to: a^2-3ab+2b^2=0 related b to a: b=(4/3)^x*a and set c=(4/3)^x so b=ac, skipping a step we get: a^2*(1-3c+2*c^2)=0 -> either a=0 which is impossible because 3 to any power will never be zero, or 1-3c+2*c^2=0 -> c=1 or 1/2 -> for c=1, (4/3)^x=1 gives x=0, for c=1/2, (4/3)^x=1/1 gives x=log base 4/3 of 1/2. * I hope I didn't make any typographical error x=log base 4/3 of 1/2=-log(2)/(2log(2)-log(3)) (any "proper" base, of course).
Given that you are elevating a root to a power, I believe you should check that your answer satisfies the original equation / makes part of the domain of the functions at both sides of the equation. That being check the value of the right side being positive and the value in the square root being greater or equal to zero
I found it to be the hardest way.... Take it easy... Square both sides first and cancel out 3^x-4^x from both sides. Then 90 % done. Thus way, we may end up with x= log(1/2) / log(4/3). Which is the same thing as log(2) / log(3/4)
How about directly... We have sqrt( 12^x-16^x ) = 3^x - 4^x Divide by 4^x - which becomes 16^x if you pull it into the squareroot... look closely, you have sqrt( (3/4)^x - 1 ) = (3/4)^x - 1 "something" equals its own squareroot?! => "something" equals either 0 or 1! (we ARE operating in a Field here!) so (3/4)^x = 1, or (3/4)^x = 2 so x = 0, or x = log base 3/4 of 2 😇
@@seanfraser3125 no problem bro, the reason you didn’t get x=0 is because when dividing by sqrt(3^x-4^x) that quantity can be 0 and that happens at x=0.
When dividing by sqrt(3^x-4^x) you should state that this quantity does not equal to zero in the resulting equation. Although it may be part of the solution in the original equation. You can get all the solutions by simply squaring both sides putting in consideration the two +- cases
Here is a thought for today: Why don't you take a few hours off, and go read the Shakespeare play Romeo and Juliet, preferably with a good glossary right there in the text so you can understand it. I write this because maybe if you see the true power and elegance of this play you won't persist in your childish and semi-irritating "or not 2b" all the time.
@@tygrataps indeed. If jxs63j practiced what he preaches, he wouldn't embarrass his good self demonstrating lack of general knowledge. BTW, jokes like this lighten the presentations and give style.
You're a good sport@@SyberMath. As a younger brother, I got plenty of grief from my sister, who was pretty much right. You'll notice I gave the wrong play in my original post, but I'm sure reading either one seriously would not hurt. They are so deep about what matters in Life besides Math that once you get the spirit you will surely recant on your attempt at this kind of "humor", and maybe even desist! And now onto today's posts!
@@JXS63J Do all Shakespeare fans make identical conclusion on "what matters in life"? Is it, or is it not, possible that some Shakespeare fans may conclude that "this kind of humour" is ok to promote? Is Shakespeare the only wise man who ever walked on this planet, for others to derive "what matters in life" from? Are you perhaps too exclusive? You may benefit from this thought: someone likes a priest, someone the priest's wife, and someone the priest's daughter!
![The most beautiful equation in math, explained visually [Euler’s Formula]](http://i.ytimg.com/vi/f8CXG7dS-D0/mqdefault.jpg)
Squaring both sides of the given equation, then dividing thru by 4^x, term-by-term, we get:
(3/4)^x - 1 = [(3/4)^x]² + 1 - 2(3/4)^x.
Letting u = (3/4)^x, and collecting all like terms to the right side of the equation, we get:
u² -3u + 2 = 0
(u-1)(u-2) = 0
so u = 1 or u = 2.
Back-sub: (3/4)^x = 1 => x = 0
and (3/4)^x = 2 => x = ln2/[ln3 - ln4].
I got 0 and log base 4/3 of 1/2...
set a=3^x and b=4^x (and not b=2^2x which is a complication as 4 is a factor of both 12 and 16).
substitute and square both sides to get:
ab-b^2=a^2-2ab+b^2, which simplifies to:
a^2-3ab+2b^2=0
related b to a: b=(4/3)^x*a and set c=(4/3)^x so b=ac, skipping a step we get:
a^2*(1-3c+2*c^2)=0 -> either a=0 which is impossible because 3 to any power will never be zero,
or 1-3c+2*c^2=0 -> c=1 or 1/2 ->
for c=1, (4/3)^x=1 gives x=0,
for c=1/2, (4/3)^x=1/1 gives x=log base 4/3 of 1/2.
* I hope I didn't make any typographical error
x=log base 4/3 of 1/2=-log(2)/(2log(2)-log(3)) (any "proper" base, of course).
You had me glued to the screen till the end, well done, loved it.
Simpler to set a=4^x and so you don't get to the 4th power and stay with squares (2nd power) and as 4 is a factor of both 12 and 16.
Substitution only make it more complicated. Just square both side, you'll get (3^x-4^x)(2*4^x-3^x)=0
By observation we can see that x must be less than or equal to 0 and right away x=0 is a solution.
Got ‘em both!
Given that you are elevating a root to a power, I believe you should check that your answer satisfies the original equation / makes part of the domain of the functions at both sides of the equation. That being check the value of the right side being positive and the value in the square root being greater or equal to zero
I found it to be the hardest way.... Take it easy... Square both sides first and cancel out 3^x-4^x from both sides. Then 90 % done. Thus way, we may end up with x= log(1/2) / log(4/3). Which is the same thing as log(2) / log(3/4)
pretty good!
How about directly...
We have sqrt( 12^x-16^x ) = 3^x - 4^x
Divide by 4^x - which becomes 16^x if you pull it into the squareroot... look closely, you have
sqrt( (3/4)^x - 1 ) = (3/4)^x - 1
"something" equals its own squareroot?! => "something" equals either 0 or 1! (we ARE operating in a Field here!)
so (3/4)^x = 1, or (3/4)^x = 2
so x = 0, or x = log base 3/4 of 2 😇
Much faster indeed. √(ab-b^2=a-b-> √b√(a-b)=a-b -> a=b - impossible; √b=√(a-b) and that's it
Yes
simple solution is assume sqrt(ab-b^2)=a-b so b(a-b)=(a-b)^2 ... b=a-b ... a=2b ... now by log we can solve it, it is simpler than your solution 😊
sqrt(12^x - 16^2)
=sqrt(4^x(3^x - 4^x))
=2^xsqrt(3^x - 4^x)
Divide both sides by sqrt(3^x - 4^x):
2^x = sqrt(3^x - 4^x)
Now square both sides:
4^x = 3^x - 4^x
-> 2*4^x = 3^x
-> 2 = (3/4)^x
So x = log(2)/log(3/4)
EDIT: x=0 is also a solution
You forgot x can be 0.
@@moeberry8226 Ah, you’re right! Thanks for pointing that out
@@seanfraser3125 no problem bro, the reason you didn’t get x=0 is because when dividing by sqrt(3^x-4^x) that quantity can be 0 and that happens at x=0.
When dividing by sqrt(3^x-4^x) you should state that this quantity does not equal to zero in the resulting equation. Although it may be part of the solution in the original equation.
You can get all the solutions by simply squaring both sides putting in consideration the two +- cases
Ln2/ln0.75
Or ln(2)/[ln(3)-2*ln(2)]
👍 like 👍
Thank you 👍
The answers I got are
x = 0
x = ln(2)/ln(3/4)
3:13 2 to the x is equal to a, not to 2...
1/_50-6*-3
Prikol
a) 3-:4-3/2-10
Sub a=3^x and b=4^x to solve the equation. 😉😉😉😉😉😉
In this case it makes more sense than 2^x
x = 0 works for me ....
Here is a thought for today: Why don't you take a few hours off, and go read the Shakespeare play Romeo and Juliet, preferably with a good glossary right there in the text so you can understand it. I write this because maybe if you see the true power and elegance of this play you won't persist in your childish and semi-irritating "or not 2b" all the time.
Isn't "to be or not to be" from Hamlet?
@@tygrataps indeed. If jxs63j practiced what he preaches, he wouldn't embarrass his good self demonstrating lack of general knowledge.
BTW, jokes like this lighten the presentations and give style.
Thank you!
You're a good sport@@SyberMath. As a younger brother, I got plenty of grief from my sister, who was pretty much right. You'll notice I gave the wrong play in my original post, but I'm sure reading either one seriously would not hurt. They are so deep about what matters in Life besides Math that once you get the spirit you will surely recant on your attempt at this kind of "humor", and maybe even desist!
And now onto today's posts!
@@JXS63J Do all Shakespeare fans make identical conclusion on "what matters in life"? Is it, or is it not, possible that some Shakespeare fans may conclude that "this kind of humour" is ok to promote?
Is Shakespeare the only wise man who ever walked on this planet, for others to derive "what matters in life" from?
Are you perhaps too exclusive?
You may benefit from this thought: someone likes a priest, someone the priest's wife, and someone the priest's daughter!