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

[xiangge]

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

[secretaryfig]

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

[danielemicucci]

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

[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! 👍👍

[Exonorm]

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

[danpul]

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

[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) ≈
OK - small difference due to rounding

[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!

[moeberry]

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.

[HoSza]

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

[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?

[oahuhawaii]

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).

[Reyansh-lcez]

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

[peace]

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

[ΓιώργοςΧαντζικος-ηπ]

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<x2
2^x1<2^x2 and 3^x1^2<3^x2^2 therefore f(x) is a strictly increasing function therefore it’s a 1 to 1 function
Show that x=1 is a solution
This works for the second one as well

I know I am late but seeing this problem as a student before and after learning more about functions made me understand that when u gain more knowledge in math the tools u gain may be very specific (or not ) but u can always create circumstances that they work in . I know it may not be part of math taught in the pre Uni school years for most of the world but here in Greece it’s the first chapter that we learn at the last year before university. Much love and support ❤

[ThePayner]

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.

[cHrtzbrg]

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

[shruggzdastr-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)?

[fatgrandpa]

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