solving a quadratic exponential equation with different bases
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- Опубліковано 28 чер 2022
- We will learn how to solve a quadratic exponential equation with different bases 2^x*3^(x^2)=6. We will use the rules of exponents, logarithm, and the factoring of a trinomial method (the tic-tac-toe method). This algebra tutorial is suitable for algebra 2 students, precalculus students or math competition students.
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#justalgebra
By a look we can say that 1 is a solution
I solve it without looking video and find second solution
@@danpul9300 O
@@tbg-brawlstars
1 is definitely a solution.
Definitively.
😋
But we should prove that 1 is the only one solution
@@grenouilles Yes
I have a simpler way: divided both sides by 6, we get 2^(x-1)*3^(x^2-1) = 1; Apply Log 3 on both sides, we get: (x-1)Log 2 + (x^2-1) = 0
We see the root of x=1 right away; then divided both side by x-1, we get x+1 = -log 2, and the second root is x=-1-log2, which is -log6
When you say “log” you mean log base 3 right?
@@omerhalaby2337 Yeah, all the logarithms taken in his situation are log base 3 in the places where the base matters.
Why is -1 - log2 = -log 6
@@orgito627 -1 = -log3
-1-log2 = -(1 + log2) = -(log3 + log2) = -log(2*3) = -log6
This works specifically because we are doing log base 3 throughout the entire solution.
Genius
2^X * 3^(X^2) = 6
LN[2^X * 3^(X^2)] = LN[6]
XLN[2] + (X^2)LN[3] = LN[6]
REARRANGE FOR QUADRATIC FORMULA, AND VOILA
Is that legal
@@Kisnowar yes
@@Kisnowar no you can get yourself arrested careful
@@abi3135 😂😂
What does LN mean?
Algebra is beautiful
Belive me it isn't, it's algebra algebra =algebra only 🙂
Nice question and excellent explanation! Two points impress me most: 1) the trick for changing different bases to the same base; 2) the trick for factoring the quadratic equation. Wonderful job! 👍👍
I used a different method!
First I wrote 3^x² as (3^x)^x and then 3^x * (3^x)^(x-1) now 2^x*3^x=6^x
Divide both sides by 6 to end up with
6^(x-1)*(3^x)^(x-1)=1 the powers are the same so we have
(6*3^x)^(x-1)=1
This is true for x-1=0 (x=1) and for 6*3^x=1
Taking ln on both sides we have ln6+x*ln3=0 which is true for x=-ln6/ln3
ln(2^x * 3^x^2) = ln(6)
x*ln(2) + x^2 * ln(3) = ln(6)
Rearrange and use quadratic formula.
You surprise me every time! 👍
Much more natural approach is to take the natural logarithm of both sides first. Works in 99% of the cases.
Yeah but here taking logarithm base 3 (instead of natural logarithm) did give a bit of an advantage.
I find solution in other way:6=2^1*3^1 so we can write this equation like 2^(x-1)*3^(x²-1)=1.Then I noticed than 3^(x²-1)=3^((x-1)*(x+1)),so we have (2*3^(x+1))^(x-1)=1.Put for both sides log with base 3 and rewrite it like (x-1)log(base 3)(2*3^(x+1))=0.First solution is 1,then find second.2*3^(x+1)=1 (6*3^x=1) x=-log(base 3)6
Could you approach the sum question by converting the 6 in terms of the x-exponent to: 6^x^0? Would that be any more helpful in finding a way to solve that equation (assuming that it has a solution)?
Looking at that equation I know that the root is going to be 1 without even doing any algebra. As a computer engineer I deal with base conversions daily as a rule. Since 2 and 3 are the only factors of 6, that means that x has to be 1. BTW - I owe my career to a math teacher. My trig and later pre-calc teacher let me hack away on the only Apple ][e in the school (this was 1986) and I taught myself assembler on that. Math teachers are the best!
I did it with ln and found
2^x × 3^x² = 6
e^(x ln2) × e^(x² ln3) = e^ln6
e^(x ln2 + x² ln3) = e^ln6
x ln2 + x² ln3 = ln 6
And with the quadratic formula, x = ( -ln2 ± √[ (ln2)² + 4 (ln3) (ln6) ] ) / ( 2 ln3 )
Was I wrong somewhere?
You are right;you have to simplify it more though
The expression under root can be written as (ln2)^+4ln3(ln2+ln3)
since ln 6=ln2+ln3
then that expression under root can be expressed as (ln2+2ln3)^2
if you sub that then you get answer as 1 and -log base 3 6
@@aks0736 thanks!
using ln 6=ln2+ln3 you can write the equation in ln2, ln3, x; divide by trivial solution (x-1) and get a quite simple second solution.
@@Bobbel888 the simple solution is obvious as 2 x 3 = 6 thereore the trivial answer has to be x = 1
thank you for such a good video
very cool video :) Thanks :)
Can u do properties of tetration
You can use Newton’s method for your bonus question. Along with derivatives and finding critical points to see when the function is increasing or decreasing and since we know it’s set to a constant function. There are only a few solutions. And the graph will not intersect the line y=6 anymore.
🤣
"just algebra"
No calculus allowed lol.
I imagine it can't be done with just algebra though, sadly.
with initial guess x=1 you get to one of the solutions x≈1.108 pretty fast, and with initial guess x=-1, you get x≈-1.251
have to use excel spreadsheet though because i aint evaluating -(2^x+3^(x^2)-6)/(ln(2)*2^x+ln(9)*3^(x^2)*x)+x by hand
Why not directly take the log of the original equation? ---> x*log 2 + x²*log 3 = log 6 which is a simple quadratic equation.
I'd like to see a complex number solved in negative base 10.
Won't it just be easier to initially 'ln' both sides? It's easier to write down and easier to use while using the quadratic formula to calculate x.
It doesn't matter what base you use, the answer you get will be the same. Having said that, I would have gone with ln too, but only because it's so prevalent throughout calculus
What is In?
@@MrSandman610 log to the base of e
@@marcushendriksen8415 yeah, I agree but it’s just easier to write down
Thank you for the nostalgia. When I was ill in elementary school I always watched the equivalent to Open University TV. It was always like watching a wizard making numbers dance. I didn't understand a thing but I couldn't stop watching.
I found the second root as : - (1 + ln(2)/ln(3))
I think it equals the solution inside the video.
I may be wrong, but I just applied the ln on both sides and the equation turns into a quadratic equation for x with real coefficients and two real distinct solutions. The first of these two solutions is 1 and the second is approx. -1.63093, which we can verify numerically as a solution.
Just take the log of both sides, reorganize, and factor to find the solutions to x.
(2^x) * 3^(x^2) = 6
x*log(2) + (x^2)*log(3) = log(6)
log(3)*(x^2) + log(2)*x - log(6) = 0
(x - 1)*(log(3)*x + log(6)) = 0
x - 1 = 0, log(3)*x + log(6) = 0
x = 1, x = -log(6)/log(3) = -log(2)/log(3) - 1
Note that -log(6)/log(3) is an easier form to compute than log3(6).
Just do 2^x *3^x² = 2*3
On comparing both sides we get 1 as answer
What about the square on top of 3 in the LHS. By comparing it you will get +1,-1
2^x . 3^x2-1 =6=2×3
(2 to power x-1 )×(3 to power x^2 -1) = 1=2 to power zero× 3to power zero
By comparing both sides
X-1=0 and x^2-1= 0
X=1 and x=+1,-1
.................x has two values ..... That is +1 and -1...... Check it if I'm wrong....
I also did the same thing but I am not sure if it is correct or not. But I think -1 will not happen. If we put -1 in LHS that will give us 3/2 that is not equal to RHS.
Ans = -[1+log(3,2)]
The second problem would be so easy if there was adding 1
Sorry, does somebody knows why if you have any number raised to log of base of the number you have as the base of the expression and the argument being the number it was before, give you the number you have before if you solve that. I'm sorry if I couln't explain me well.
What about this one?
(2^x)(3^x)=6
(2^x)(3^x)=(2¹)(3¹)
Since bases are equal power must be equal too
Therefore (i) equ.----->
(2)^x=(2)¹
x=1
(ii) equ.------>
(3)^x^2=(3)¹
x^2=1
x=√1
x=1
Therefore x=1 for both cases
That doesn't find all solutions though.
x²-1 = 0
x=1 is easy to see, so that is definitely a solution, but it's not the only solution.
If you only needed one solution, however, that would indeed be the fastest one to find.
Nice
Some number theory pls...
The simple asner is x = 1 the more complex answer aproximates -1.630929 if you want it more precise do it yourself
How to solve X in (e^aX) + (e^bX) = c ? , where a,b,c is a real number
We know that 6 = 2*3, then
2^x*3^{x^2) = 6 = 2*3
2^x = 2 and 3^{x^2) = 3
Here
2^x = 2, x = 1
3^{x^2) =b3, x = 1 or x = -1
Then the answers is x = 1
All though I liked the solution isn’t it better to say :
Let f(x)=2^x times 3^x^2
Prove f(x) is a 1 to 1 function therefore it has only one real solution like this
For every x1,x2 that exist in the definition set ( which is R ) , with x1
Got log18(6)
Hey quick question if you sub x = 1 wouldn’t it be 2•3^2 = 2•9 = 18
It is not 3^2 but 3^1 because it is 3^(1^2)
2^1 × 3^1^2 =6 x=1
Does this complicating the equation
U can to find it in a quite simple way:
2^(x).3^(x²)=6
2^(x).3^(x²)=2¹.3¹
X=1
So easy
6=3*2 sow we eliminate the two bases and we get x cube =1 which is x=1
Can't take the log?
x² ln(3) - x ln(2) - (ln(2) + ln(3)) = 0?
Is that allowed?
Okay yeah same idea as other people
x=1
2
Solution by insight
2×3=6
x=1
How do you know that
-1 + log(3,6) = log(3,6)-log(3,3)
1 can be written as log(a,a).In this case, it's -1 so -log(3,3)
@@bprpmathbasics i dont understand the factoring. If we multiply it back we get x^2 and -log3(6), but for the middle part i get x*log3(6) - log3(6). I do not understand what the -1 + log3(6) = log3(2) helps, because there is no -1 in a sum but there is an x * log3(6) in the product i get.
What do I not understand?
@@RandomBW I do think another way
x²+xlog(3,2)-log(3,6)=0
log(3,6)=log(3,6)*1=log(3,6)*log(3,3)
Since log(3,2)=log(3,6/3)=log(3,6)-log(3,3)
Looks familiar?
Cause it is form of a+b and ab from equation (x+a)(x+b)=0
(x+log(3,6))(x-log(3,3))=0
(x+log(3,6))(x-1)=0
Hope this one is not quite circular than before
@@RandomBW And he ll never replay to that :)
X 1
can we do
ln(2^x3^x^2) = ln(6)
ln(2^x) + ln(3^x^2) = ln(6)
xln(2)+x^2ln(3) =ln(16)
x²ln(3) + xln(2) - ln(6) = 0
then use quadratic formula
Oo nice problem!!!
12X = 300.
(In bowling)
Why not simply take the log of the original equation (in any base) and solve the resulting quadratic equation?
Ok first of all what is a log and where it come from? What would 1 do? And why is it zero? I mean how does X with a little 2 on top plus Xlogwith a little 3 on bottom with a little 2 on top minus a X with a 3 on bottom with a 6 equal a 0? Where does this stuff come from?
there are two solution
(-1-sqrt(5))/2 and (-1+sqrt(5))/2)
It's ok like like u flip the whole equation from the first equation.thats cool. Please correct me please thank you
Soo.. . holding plush PokeBall boosts your math proficiency. Got it!
X = 1 is obvious just looking at the video thumbnail.
Vous avez compliqué les étapes il suffit de décomposer le chiffre 6 en 3&2"
i actually did 3x^2 - 2x and got 1
I actually got 1 by doing it in my mind;-;
Извините , я - по-русски . Обе части уравнения >0 . Логарифмируем обе части. Получаем квадратное уравнение : lg(3)*x^2+lg(2)*x-lg(6)=0 . X1=1 ; X2=-lg(6)/lg(2) . И ВСЁ!! С уважением,lidiy27041943
X = 1
x = 1 is a trivial but valid solution
Cool
1
answer is 1 and 1/2
i figured this out in about 10 seconds lmao
1啊
2ⁿ×2ⁿ^² = 6
2ⁿ×2ⁿ^² = 2×3
Entonces n=1
Х=1
x=1!!!
x = x^2
x^2 - x = 0
x ( x - 1 ) = 0
x = 0 or x = 1
check : 2^0 * 3^0 = 6
1 = 6 ( false )
2^1 * 3^(1^2 ) = 6
2 * 3 = 6
6 = 6 ( true )
Therefore, x = 1 is the solution.
Where does the assumption x = x^2 come from?
@@cosmicvoidtree 2^x × 3^x^2 = 6
2^x × 3^x^2 = 2 × 3
2^x × 3^x^2 = 2^1 × 3^1
So for 2^x = 2^1, x = 1
And 3^x^2 = 3^1
So x^2 = 1 => x =1/(-1)
We already know x=1, and now x²=1/-1
But since there are 2 actions being taken i.e negative × positive, answer must come out negative so therefore solution -1 is rejected
So we get x=1,x²=1(only solution)
So x=x² and x=1/x²=1 is a solution
You missed a solution though
Well first of all it’s 1, but that’s not very interesting
xln2+x^2*ln3=ln6
ln3*x^2+ln2*x-ln6=0
You bet your sweet ass im using the quadratic formula
x=[-ln2+sqrt(ln2*ln2+4*ln3*ln6)]/2ln3
I’m still mad that people say this notation is wrong
[-ln2-sqrt(ln2*ln2+4*ln3*ln6)]/2ln3
Oh also ln spits out infinite values, but Im tired so i wont work out how that affects the answer
That Pokéball in his hands is cute.
I did something wrong but i dont know what :(
2^x *3^x^2 = 6
2^x * 3^2^x = 6
2^x * 9^x = 6
18^x = 6
log(6)/log(18) is proximately = 0.62
3^x^2 isn't equal to 3^2^x
@@joaoferroviarias8919 but it doesnt matter how you change the exponents as a exponent rule, i thaught?
@@uchihaitachi6852 that would only be the case if the exponent operates on the entire thing, like in (3^x) ² but here the 2 acts as an exponent only on the x
@@NeelTigers ak ok that makes sense
@@NeelTigers how to know if 2 act as the exponent on x only, and not the entire thing?
Can someone explain me why is he always holding a pokeball? 😂 I'm new to this channel
It's the mic.
RHS is positive, take the log, rest us trivial.
For the second question: my answer became +- the square root of (log(4)/log(3)). And I rearranged a little in the beginning and then used rule of logrithms. But when I plug in the answer to test it, it only moves closer and closer to 6,17849, and not 6, why is that?!
And I double checked on Photomath, it does not show how to solve it, but on the graph I can roughly see that it is the right answer?
If you are talking about the second question, could you explain it more detailed, please?
Can't we just
2x*3x²=6
so 2x*3x²=6x³=6
So x³=1
Then x=1
bruh bruh... I guess you were either sleepy or high. Sorry if I sound rude, but the thing is that x is in exponent form and not a part of base to perform simple algebra.
Just take ln of both sides first
∞
It’s easy answer is 1
Just take natural logarithm on both sides and solve
Reminds of an IQ problem to check your insight, and how long it takes to give an answer of 1.
I took the first glance at it and immediately thought that the answer was 1. But then reconsidered, took another closer look and though the answer was log(18,6). But the video and the conment section said that the answer was 1 and -log(3,6).
I thought that 2 was supposed to act as exponent over the entire 3^x, not just x alone. I guess I need to learn logarithmics again.
👍👏
Obviously x=1.
My guy you only qualify to teach intelligent people 😌
Just take log of both sides:
log(2^x * 3^x²) = log(6)
x log 2 + x² log 3 = log 6
x² log 3 + x log 2 − log 6 = 0
Now we have a quadratic equation, so we can solve by factoring or using quadratic formula:
x² log 3 + x (log 6 − log 3) − log 6 = 0
(x² log 3 − x log 3) + (x log 6 − log 6) = 0
x log 3 (x − 1) + log 6 (x − 1) = 0
(x − 1) (x log 3 + log 6) = 0
x = 1 or x = −(log 6)/(log 3) ≈ −1.63093
Check:
2^(1) * 3^(1²) = 2 * 3 = 6
2^(−1.63093) * 3^((−1.63093)²) ≈ 6.000004
OK - small difference due to rounding
Bro x=1
Ummmm, x=1
2^x * 3^x^2 = 6
2^x * 3^x^2 = 2*3
if bases is equals Then exponents is equals too
x + x^2 = 1 + 1
x^2 + x - 2 = 0
x1=1
x2=-2
-2 is not a solution
@@musicismylife6172, are you sure about that?
Everybody knows, but i know
BPRP: We can divide 2 to the x. That'll give us 6 times...
Me: NO HAHA YOU FUCKED UP
BPRP: 2 to the negative x
Me:....oh...yeah....
I don't think that the bonus question has a nice answer either. But I didn't devote any substancial amount of brain juice into it.
My School hasn't taught us log yet so if we go by our school's method it would be like this
2^x • 3^(x^2) = 2• 3
Therefore, 2^x = 2^1, x=1
3^(x^2)=3^1,x^2=1
(x^2)-1=0, (x+1)(x-1)=0
Since 2^x = 2^1, x= -1 is not a sol
Therefore, x=1
(Note: my year hasn't even learnt quadratic formula or the plus or minus sign yet, usually we won't even bother proving x is not equal to -1)
Interesting. What grade are you in?
@@kanna-chan6680 My country doesn't have the concept of 'grades' as it's a SEA country, Malaysia. Even though most western ppl tend to think we asians are better in studies, thats not always the case lol. I'm 15 and in secondary school, form 3. For secondary school, there are pre-secondary(we call peralihan),Form 1,2,3,4,5,6. Pre-secondary is for those who failed Malay(or maybe other main subjects,is around 1 year and if u pass u get into secondary).The other forms are normal but form 6 is pre-college/university. I study at half-public school, a private chinese school here would have a much higher difficulty and more advanced
after watching tumbnail. i thought for a moment and answer is x=1
Yeah, it's 1 isn't it?
0:42 "still can not do anything"... "can't take log". Huh? Why can't you take ln of both sides, bro?
Fundamental theorem on Integers says Every positive integer can be written exactly in one way as a product of Primes
( except for different placements) So write the right side as 2.3 and equate the powers to get x=1 from both.
Also the question is meaningless if the two values of x from equating the exponents of 2and 3 yield two different
solutions to draw the conclusion that the an is inconsistent.
if you are solving in integers, then you are right, but, as you can see, there is also a non integer solution
Anh why he holding pokemon ball
😂😂😂
That gives him superpower of remembering things
Hey bprp i solved the given question of yours equation.2^x+3^x^2=6.
The solution is [0.86328125, 0.875].hope you will notice this comment