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.
Get the "algebra" t-shirt 👉 blackpenredpen...
Use "WELCOME10" for 10% off
-----------------------------
If you find my channel helpful and would like to support it 💪, then you can
support me on Patreon: 👉 / blackpenredpen
or buy a math shirt or a hoodie: 👉 bit.ly/bprp_merch
-----------------------------
"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!
Subscribe for future algebra tutorials 👉 bit.ly/just_al...
-----------------------------
#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

[@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.

[@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

[@MarieAnne.]

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

[@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).

[@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

[@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.

[@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.

[@ΓιώργοςΧαντζικος-η5π]

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

[@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

[@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

[@davidmurphy563]

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

[@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?

[@gamerdudestube838]

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

[@anestismoutafidis4575]

2^1 × 3^1^2 =6 x=1

[@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.

[@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

[@ciberiada01]

You surprise me every time! 👍

[@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"

[@alfahentriza5571]

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

[@simonmitchell6516]

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

[@Wolkenphoenix]

very cool video :) Thanks :)

[@AndVer]

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

[@srilatapn6367]

X 1

[@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

[@에스피-h8t]

Solution by insight 2×3=6 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

[@dibyojyotibhattacherjee897]

Some number theory pls...

[@akshayakumarmalviya4749]

answer is 1 and 1/2

[@RadekBuczkowski-h2y]

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)

[@yaseenelhosseiny]

X=1

[@lshowt]

1啊

[@nhtaee5693]

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

[@sikosiko_tinpo_gansha.gansha]

x=1!!!

[@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"

[@girishjayansenthilkumar4716]

That Pokéball in his hands is cute.

[@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.

[@coc0a_mst]

That sounds more like a you problem m8

[@johnporcella2375]

@@coc0a_mst 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.

[@gamingbutnotreally6077]

Just take ln of both sides first

[@tv-ll5vj]

Everybody knows, but i know

[@SyberMath]

I love this problem! Nice work!!! 😍 I made a video response. Let me know what you think 😊

[@SyberMath]

@@bprpmathbasics Thank you, professor! You've got great content and a very nice way of explaining things. Hoping to meet you one day 💖

[@jasonkara7281]

Ummmm, x=1

[@juandelacruz9027]

There is an easier and simpler way. (2expx)(3expsquared) = 6. Factor 6 into (2)(3). compared both sides of the equation noting their respective exponents, you will have x= 1 and x squared = 1. Solving we have x = 1 and x = plus or minus 1. We accept only the value 1 and discard -1.

[@oahuhawaii2141]

But you are missing the other solution of -log(6)/log(3) or -log(2)/log(3)-1 .

[@juandelacruz9027]

@@oahuhawaii2141 to each his own

[@oahuhawaii2141]

@@juandelacruz9027: Weak excuse.

[@juandelacruz9027]

​@@oahuhawaii2141 Oh i'm sorry strong strong strong great great mathematician of the multiverse of everything! We are not worthy! we are not worthy!..................hooohahhh!!!!

[@siddhjain3075]

I solved it within seconds by comparing both sides...... As it can be written as 2^x . 3^x^2 = 2^1 . 3^1

[@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.

[@eldorado318]

Obviously x=1.

[@terranceparsons5185]

Yeah, it's 1 isn't it?

[@marshalls36]

好，奖 1 万元

[@thegiant080]

Anh why he holding pokemon ball

[@hasnain9654]

😂😂😂

[@hasnain9654]

That gives him superpower of remembering things

[@albertnawab9211]

Yeah, wheb we see, we knew that x= 1 Why you make it confused with logx? In daily life, nobody use this calculations.

[@vjlaxmanan6965]

Stupidly complicated soln :(

[@du42bz]

I got something wild from trying the route through the quadratic formula x² + x*log(3,2) - log(3, 6) (x + (log(3,2)/2)²)² - log(3, 6) - (log(3,2)/2)² x = √((log(3,2)/2)² + log(3,6)) - (log(3,2)/2)² ≠ solution of the video Something somewhere must have seriously gone wrong

[@plislegalineu3005]

use log property log₃ (6) = log₃ (3 × 2) = log₃ (3) + log₃ (2) = 1 + log₃ (2) Then factor the inside of the square root as a perfect square

[@chitlitlah]

Thanks KZbin. I worked on this for about an hour last night before solving it, posted a reply, and they marked it as spam so only I can see it. Unbelievable...

[@chitlitlah]

I'm going to copy and paste my initial reply but just a few lines per post and see if it stays up: It may be the way it has to be typed out, but I don't know how you went from the first step to the second step or why. The first one is already in the form ax^2 + bx +c and it's equal to 0. So here's what I got using the quadratic formula. Assume any log function is base 3; it looks neater if I don't explicitly put it.

[@chitlitlah]

a = 1, b = log 2, c = - log 6 x = (-log 2 +/- sqrt((log 2)^2 + 4 log 6)) / 2 I eventually figured out that log 6 is equal to log 2 + 1 which helps us get:

[@chitlitlah]

Let me throw some words in here so KZbin won't think it's spam... x = (-log 2 +/- sqrt((log 2)^2 + 4 log 2 + 4)) / 2 (log 2)^2 +4 log 2 + 4 = (log 2 + 2)^2 x = (-log 2 +/- (log 2 + 2)) / 2

[@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?

[@nusretgulmammadov5144]

x=1

[@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

[@rubiks6]

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

[@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

[@Reyansh-lc8ez]

2^x . 3^x = 2.3 x=1 (I am an 8th grader)

[@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.

[@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

[@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.

[@schizoframia4874]

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

[@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!

[@maydin34]

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

[@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

[@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)?

[@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

[@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

[@soniaalboresi5488]

x=1

[@AzeOfSpadez]

i figured this out in about 10 seconds lmao

[@dynamic-gaming1879]

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

[@angirasnazar59]

Oo nice problem!!!

[@zekeriasvarg530]

I want it solved without the log methods!

[@MathwithLukgaf]

Nice

[@ridosaputra5273]

Why dont tranform 6 into 2×3?

[@sifisomavimbela8838]

My guy you only qualify to teach intelligent people 😌

[@avspranavchowdary2230]

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

[@mel_ha22]

I didn't understand anything T_T

[@nestormanuelrussibogota2554]

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

[@theuserings]

Can u do properties of tetration

[@OmarJIBAR]

Cool

[@melk33]

Yes why. Why do we need wasting time

[@TheBatugan77]

12X = 300. (In bowling)

[@arsss967]

It’s easy answer is 1

[@arraser84]

Got log18(6)

[@lavanyamendiratta2735]

Bro x=1

[@ЛидийКлещельский-ь3х]

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

[@patrickmcbrearty4579]

X = 1

[@BerndReinhardt]

X=1

[@equinoxe225]

X=1

[@michaelpenklis7580]

∞

[@heisenberg4330]

1

[@ИгорьАрсоу]

Х=1

[@0011peace]

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

[@xerveschex5761]

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

[@byronwatkins2565]

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

[@HoSza1]

0:42 "still can not do anything"... "can't take log". Huh? Why can't you take ln of both sides, bro?