Solving an Exponential Equation

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[xXJFARGAMERXx]

2^x² = 3^x
x² = log₂(3^x)
x² = xlog₂(3)
Assuming x ≠ 0
x = log₂(3)

Let's check if x = 0 is also a solution

2^(0)² = 3^(0)
2^0 = 1
1 = 1
Yup, so the solutions are:
x = 0, or x = log₂(3)

[GourangaPL]

shouldn't we use change of base formula in the end to write it as log_2 (3) ?

[Blaqjaqshellaq]

And x^2*ln(2) - x^ln(3)=0 is a quadratic equation with roots 0 and ln(3)/ln(2).

[netanelkomm]

how about ln both sides, move everything to one side u get:

ln (2^(x^2)) - ln(3^x) = 0

move the exponents down and take a common factor x out:

x (x•ln2 - ln3) = 0

x = 0 is one solution.

then:

x•ln2 = ln3
x = ln3 / ln2 is second solution


this way you dont have to assum x != 0

[sebastienlecmpte]

Why are people always using natural log instead of the easier base?

In this cas just log base 2 on both sides. No need for power rule or anything, straight to the answer.

x= log base 2 of 3

[jeremycole]

What about something like:



√2 is a solution but WolframAlpha also gives 3.13556…as a second solution. How do you find this second solution?
Thanks.

[gdmathguy]

I am so confused! Why can you cross out the x's when it would yield different results when inputted the same value in the non crossed out x and x crossed out equasions?

[Bjowolf]

Or to make it simpler:

x = ln 3 / ln 2 = log2 (3) 😉

[alisarvar]

You made a spelling mistake. It is 2^x^x not 2^x^2

[astitva]

Hi, A small doubt...
In the question, it is given that, (2^x)^2=3^x
And when I solve it: (2^2)^x=3^x, which is 4^x=3^x and from it I can say that x=0 (a soln of it.)
But in the soln, you have written, (2^x)^x=3^x how? But in the ques, a diff eq is given, And how x=ln3/ln2 is coming?
Hopefully you will give the ans,

[kuriana]

I am so sorry. I can't understand. If you could elaborate

[neologicalgamer]

Brozowski, ln3/ln2=ln(3-2)=ln(1)=0 💀
That ain't another solution, just another way if writing "zero"

[sigmascrub]

It's been a while but iirc ln(3)/ln(2) = ln(3-2) = ln(1) = 0, no?