solving a quadratic exponential equation with different bases

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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"Just Algebra" (by blackpenredpen) is dedicated to helping middle school, high school, and community college students who need to learn algebra. Topics include how to solve various equations (linear equations, quadratic equations, square root equations, rational equations, exponential equations, logarithmic equations, and more), factoring techniques, word problems, functions, graphs, Pythagorean Theorem, and more. We will also cover standardized test problems such as the SAT. Feel free to leave your questions in the comment!
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#justalgebra

[@tbg-brawlstars]

By a look we can say that 1 is a solution

[@danpul9300]

I solve it without looking video and find second solution

[@tbg-brawlstars]

@@danpul9300 O

[@PercivalBlakeney]

@@tbg-brawlstars 1 is definitely a solution. Definitively. 😋

[@grenouilles]

But we should prove that 1 is the only one solution

[@tbg-brawlstars]

@@grenouilles Yes

[@xiangge6374]

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

[@omerhalaby2337]

When you say “log” you mean log base 3 right?

[@paulstelian97]

@@omerhalaby2337 Yeah, all the logarithms taken in his situation are log base 3 in the places where the base matters.

[@orgito627]

Why is -1 - log2 = -log 6

[@paulstelian97]

@@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.

[@steabestesnathan662]

Genius

[@secretaryfig5364]

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

[@Kisnowar]

Is that legal

[@Amoeby]

@@Kisnowar yes

[@abi3135]

@@Kisnowar no you can get yourself arrested careful

[@scipionoir9660]

@@abi3135 😂😂

[@MrSandman610]

What does LN mean?

[@DrLiangMath]

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! 👍👍

[@fish8622]

Algebra is beautiful

[@billielish897]

Belive me it isn't, it's algebra algebra =algebra only 🙂

[@Exonorm27]

ln(2^x * 3^x^2) = ln(6) x*ln(2) + x^2 * ln(3) = ln(6) Rearrange and use quadratic formula.

[@danielemicucci4788]

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

[@danpul9300]

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

[@BitwiseMobile]

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!

[@ciberiada01]

You surprise me every time! 👍

[@HoSza1]

Much more natural approach is to take the natural logarithm of both sides first. Works in 99% of the cases.

[@paulstelian97]

Yeah but here taking logarithm base 3 (instead of natural logarithm) did give a bit of an advantage.

[@mg7094]

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.

[@mynameisgood1637]

thank you for such a good video

[@Wolkenphoenix]

very cool video :) Thanks :)

[@shruggzdastr8-facedclown]

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)?

[@Hippolyte_Pequeux]

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?

[@aks0736]

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

[@Hippolyte_Pequeux]

@@aks0736 thanks!

[@Bobbel888]

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.

[@0011peace]

@@Bobbel888 the simple solution is obvious as 2 x 3 = 6 thereore the trivial answer has to be x = 1

[@moeberry8226]

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.

[@trivikram4962]

🤣

[@epsi]

"just algebra" No calculus allowed lol. I imagine it can't be done with just algebra though, sadly.

[@Samir-zb3xk]

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

[@cH3rtzb3rg]

Why not directly take the log of the original equation? ---> x*log 2 + x²*log 3 = log 6 which is a simple quadratic equation.

[@IsaacTorresProf]

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.

[@theuserings]

Can u do properties of tetration

[@oahuhawaii2141]

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).

[@davidmurphy563]

I'd like to see a complex number solved in negative base 10.

[@erixmonteza3130]

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.

[@maydin34]

I found the second root as : - (1 + ln(2)/ln(3)) I think it equals the solution inside the video.

[@ThePayner11]

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.

[@marcushendriksen8415]

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

[@MrSandman610]

What is In?

[@ThePayner11]

@@MrSandman610 log to the base of e

[@ThePayner11]

@@marcushendriksen8415 yeah, I agree but it’s just easier to write down

[@JasimGamer]

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

[@fatgrandpa9376]

Just do 2^x *3^x² = 2*3 On comparing both sides we get 1 as answer

[@bruh07271]

What about the square on top of 3 in the LHS. By comparing it you will get +1,-1

[@schizoframia4874]

The second problem would be so easy if there was adding 1

[@alfahentriza5571]

How to solve X in (e^aX) + (e^bX) = c ? , where a,b,c is a real number

[@0011peace]

The simple asner is x = 1 the more complex answer aproximates -1.630929 if you want it more precise do it yourself

[@Nil_11186]

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

[@epsi]

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.

[@sk_____007]

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....

[@bruh07271]

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.

[@angirasnazar59]

Oo nice problem!!!

[@MathwithLukgaf]

Nice

[@sie_khoentjoeng4886]

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

[@dibyojyotibhattacherjee897]

Some number theory pls...

[@byronwatkins2565]

Why not simply take the log of the original equation (in any base) and solve the resulting quadratic equation?

[@simonmitchell6516]

Ans = -[1+log(3,2)]

[@txikitofandango]

Can't take the log? x² ln(3) - x ln(2) - (ln(2) + ln(3)) = 0? Is that allowed?

[@txikitofandango]

Okay yeah same idea as other people

[@hguy4100]

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

[@xerveschex5761]

Soo.. . holding plush PokeBall boosts your math proficiency. Got it!

[@Wolfie_Desu]

Hey quick question if you sub x = 1 wouldn’t it be 2•3^2 = 2•9 = 18

[@Nzonzimi]

It is not 3^2 but 3^1 because it is 3^(1^2)

[@AndVer]

there are two solution (-1-sqrt(5))/2 and (-1+sqrt(5))/2)

[@user-yg3tz8rb8q]

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

[@arraser84]

Got log18(6)

[@gamerdudestube838]

6=3*2 sow we eliminate the two bases and we get x cube =1 which is x=1

[@shemiahwalker]

It's ok like like u flip the whole equation from the first equation.thats cool. Please correct me please thank you

[@philiplauren7024]

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?

[@paunb8550]

If you are talking about the second question, could you explain it more detailed, please?

[@user-dv8gv3hu4t]

Solution by insight 2×3=6 x=1

[@nhtaee5693]

i actually did 3x^2 - 2x and got 1

[@du42bz]

How do you know that -1 + log(3,6) = log(3,6)-log(3,3)

[@mitthrawnuruodo2880]

1 can be written as log(a,a).In this case, it's -1 so -log(3,3)

[@RandomBW]

@@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?

[@jofx4051]

@@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

[@davyz9143]

@@RandomBW And he ll never replay to that :)

[@nadagigi5130]

Vous avez compliqué les étapes il suffit de décomposer le chiffre 6 en 3&2"

[@AzeOfSpadez]

i figured this out in about 10 seconds lmao

[@rubiks6]

X = 1 is obvious just looking at the video thumbnail.

[@avspranavchowdary2230]

I actually got 1 by doing it in my mind;-;

[@niceguy999918]

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?

[@OmarJIBAR]

Cool

[@anestismoutafidis4575]

2^1 × 3^1^2 =6 x=1

[@TheBatugan77]

12X = 300. (In bowling)

[@riyaziyyat1903]

👍👏

[@starpawsy]

x = 1 is a trivial but valid solution

[@soniaalboresi5488]

x=1

[@srilatapn6367]

X 1

[@willie333b]

2

[@user-pd7js7cy9m]

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

[@yiutungwong315]

X = 1

[@akshayakumarmalviya4749]

answer is 1 and 1/2

[@lshowt]

1啊

[@girishjayansenthilkumar4716]

That Pokéball in his hands is cute.

[@uchihaitachi6852]

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

[@joaoferroviarias8919]

3^x^2 isn't equal to 3^2^x

[@uchihaitachi6852]

@@joaoferroviarias8919 but it doesnt matter how you change the exponents as a exponent rule, i thaught?

[@NeelTigers]

@@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

[@uchihaitachi6852]

@@NeelTigers ak ok that makes sense

[@subliminalfalllenangel2108]

@@NeelTigers how to know if 2 act as the exponent on x only, and not the entire thing?

[@sikosiko_tinpo_gansha.gansha]

x=1!!!

[@nestormanuelrussibogota2554]

2ⁿ×2ⁿ^² = 6 2ⁿ×2ⁿ^² = 2×3 Entonces n=1

[@gabrielgomes1618]

Can someone explain me why is he always holding a pokeball? 😂 I'm new to this channel

[@oenrn]

It's the mic.

[@michaelempeigne3519]

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.

[@cosmicvoidtree]

Where does the assumption x = x^2 come from?

[@anandupadhyay5479]

@@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

[@Johnny-tw5pr]

You missed a solution though

[@dynamic-gaming1879]

after watching tumbnail. i thought for a moment and answer is x=1

[@heisenberg4330]

1

[@mellow-jello]

Reminds of an IQ problem to check your insight, and how long it takes to give an answer of 1.

[@subliminalfalllenangel2108]

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.

[@gamingbutnotreally6077]

Just take ln of both sides first

[@cmilkau]

RHS is positive, take the log, rest us trivial.

[@alielhajj7769]

Just take natural logarithm on both sides and solve

[@user-sy3qx5qv2w]

Х=1

[@ieatgarbage8771]

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

[@sifisomavimbela8838]

My guy you only qualify to teach intelligent people 😌

[@arsss967]

It’s easy answer is 1

[@deadkiller4129]

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)

[@kanna-chan6680]

Interesting. What grade are you in?

[@deadkiller4129]

@@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

[@michaelpenklis7580]

∞

[@keshavrateria]

Can't we just 2x*3x²=6 so 2x*3x²=6x³=6 So x³=1 Then x=1

[@Bhuvan_MS]

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.

[@tv-ll5vj]

Everybody knows, but i know

[@deleted..214]

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

[@eldorado318]

Obviously x=1.

[@lavanyamendiratta2735]

Bro x=1

[@johnporcella2375]

That was incomprehensible to anybody who has forgotten the rules of logarithms! See how Pre-Math channel always states what the rules are when he brings techniques in.

[@tahermahi]

That sounds more like a you problem m8

[@johnporcella2375]

@@tahermahi The presenter clearly knew the rules of logs...would it have killed him to explain them? No, not really! As I wrote, it is years since I studied logs and it would not have cost the host to summarise the general rules for the ones used. The first rule of teaching...assume nothing. The chap who presents Pre-maths would have taken a moment to remind people.

[@thefek]

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....

[@ivocosmi5303]

Veo que se complico, yo lo resolvía así: 2^x . 3^x^2 = 2^1 . 3^1 ; igualando bases idénticas obtenía: x=1 , x^2=1 ; solucionando las 2 ecuaciones obtenía: x=1 y x=-1 ; pero como solo cumple para 1 en la ec. original, entonces la solución es: x=1 "LISTO SOLUCIONADO"

[@user-ie4oz4hu6x]

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

[@musicismylife6172]

-2 is not a solution

[@user-ie4oz4hu6x]

@@musicismylife6172, are you sure about that?

[@venkataramannarayanaswamy2833]

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.

[@comcomrade6547]

if you are solving in integers, then you are right, but, as you can see, there is also a non integer solution

[@thegiant080]

Anh why he holding pokemon ball

[@hasnain9654]

😂😂😂

[@hasnain9654]

That gives him superpower of remembering things

[@jasonkara7281]

Ummmm, x=1

[@marshalls36]

好，奖 1 万元

[@user-pu5dd8tf7c]

There is a slighlty simpler solution without using a quadratic polynomial. 3^(x^2) * 2^(x) = 3^1 * 2^1 | / (3 * 2) 3^(x^2 - 1) * 2^(x - 1) = 1 | * 2^[-(x - 1)] 3^[(x - 1)(x + 1)] = 2^[-(x - 1)] | ln (x - 1)(x + 1)ln(3) = -(x - 1)ln(2) | / (x - 1) => assuming x != 1 1) x = 1 (valid solution) 2) x != 1 (x + 1)ln(3) = -ln(2) x = -ln(2)/ln(3) - 1 This is equivalent to the solution in the video: x = -ln(2)/ln(3) - ln(3)/ln(3) x = -[ln(2) + ln(3)]/ln(3) x = -ln(2*3)/ln(3) x = -log3(6)