A Nice Exponential Equation Suggested by Richiez111

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[pwmiles]

Nice problem, I had to graph it, but... with hindsight you can square-root both sides to get
+/-x = 4^x
The positive sign doesn't give any real solutions. With the negative sign, the left hand side is decreasing from +infinity to -infinity and the right hand side is increasing from 0+ to +infinity. So the graphs have to cross once i.e. at x = -1/2

[gastonsolaril.]

I solved it through Lambert W function and I got the same answer!!! "x = -1/2"
There's one other complex solution that's "-1.13 + 0.08 i"

[goldfing]

g(x) = x^2
h(x) = 16^x
A quick glance at the graphs shows that there cannot be any positive solution, because the exponential function grows much faster than the quadratic function. We must search for a negative solution, which must lie between -1 and 0, because we have a "polarity switch" of g and h between these two x values:
g(-1) = (-1)^2 = 1 > 1/16 = 16^(-1) = h(-1)
but
g(0) = 0^2 = 0 < 1 = 16^0 = h(0)
Let us try the mid (as we would do in bisection method anyway), x = -1/2.
g(-1/2) = (-1/2)^2 = 1/4
and
h(-1/2) = 16^(-1/2) = 1/16^(1/2) = 1/sqrt(16) = 1/4 as well.
Bingo, solution found.

[Paul-]

I took the ln of both sides, didn’t see an obvious path to a solution, played around with likely solutions, and came up with -1/2. Then I watched the video and saw him solve it the same way. 🙃

[rickdesper]

Not a big fan of just throwing algebraic manipulations at this equation. I prefer looking at the two functions separately to get a broad view of what's going on, and then doing the algebra. In this case, we see that, at x=0, x^2 < 16^x. Also, the lower function is a polynomial while the higher function is an exponential with a large base. So x^2 < 16^x for all positive values of x.
The next question is: are there any solutions at all? Well, if we look at negative values of x, then as x goes to negative infinity, the exponential function goes to zero, but the polynomial goes to positive infinity. So at some point they must cross. (That's over kill, already at x=-1, (-1)^2 = 1 > 16^(-1) = 1/16. )
So, once we know that we're looking at negative values of x for a solution, we don't have to chase the rabbit using ln x out of the gate. We could start by using ln (x^2) = 2 ln (-x), conditioned on x< 0. And then the rest of this explanation falls in place. My point is that we should know what values of x we are looking at before we apply the natural log function.
Or, we could take a shortcut and guess that the answer probably is a power of 2. And between -1 and 0. The first thing to check would be x = -1/2, and we're done.

[Ssilki_V_Profile]

I already thought that you forgot about the even degree, and about how to logarithm it correctly;)

I propose a problem:
f(x) = x^2
n(x) = a^x, where a>1

With negative "x", the functions intersect only once.
But with positive "x", it is possible that they will intersect twice already, for example, with "a=2",
or not intersect at all, for example, with "a=e"

it is necessary to find such "a" at which the functions touch at one point, and also to find this point.

Experimentally, I determined that "a" should be greater than 2 and less than 2.1.

[mehrdadbasiri]

Perfect...👍👍👍.
Just i have a question...
I'm trying to make the powers the same...
Something like this:
a^b=b^a
Or
a^a=b^b
Can we make like this and answer it?

[darksamoth]

i'm very interested in the equation 1/x^1/x, could you do a video on it ? i thought about trying to find the solution of 1/x^1/x=1/b^1/b and also if there was more than 2 and 4 being the only integers solutions

[josephatjose]

Thank you, brother, for a splendid solution. I wonder how to solve logx^2=×/25

[GreenMeansGOF]

2ln(x)=ln(16)x=ln(16)e^ln(x)

iff -ln(x)*e^(-ln(x))=-ln(4)

iff -ln(x)=W(-ln(4))

iff x=e^(-W(-ln(4)))

Not sure if there is a domain error somewhere nor which branch gives us the real solution.

[Mum-mum-mum-mah]

I feel bad for not solving this problem. When I go to engineering. 😢

[alextang]

This problem is SyberMath favour becaue he always uses monotonic behavior to solve this problem.
In short, x^2 increases slower than 16^x after a value. So can easily obtain one solution. For the smaller solution, it is not difficult to find because the behavior of x^2 and 16^x is reversed. 😋😋😋😋😋😋

[math-physics]

Please can you solve this cracking equations. It seems easy but hard.
log(x) + log(y) = 1
(log(x))^2 + (log(y))^2 = 2
log(x) - log(y) + (log(x))^log(y) = 3
x = ?
y = ?

[aaronsterlind]

Okay, I'm a high school drop out, but, why is X not 16 then? 16 X 16 = 16 X 16? Does it not? Now, I haven't watched the video but the equation drew me in, I checked the comments, I see many things, but not 16. Wait, two minutes went by 16 X 16 isn't the same as 16 to the 16th power, okay, my bad.

[gelbkehlchen]

Solution:
x² = 16^x ⟹
x² = 4^(2*x) |/4^(2*x) ⟹
x²/4^(2*x) = 1 = (-1/2)²/4^[2*(-1/2)]
|the same operations are done with x on the left side as with -1/2 on the right side of the equation, therefore ⟹ x = -1/2