A Homemade Exponential Equation

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(x^2-9)^x=(x+3)^x
#ChallengingMathProblems #ExponentialEquations #Exponentials
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[YahiaNebti]

I don't think that x=-3 is a valid solution because you'll get : 0^-3 = 0^-3
And you can't raise zero to a negative power or to 0 power because 0^-3 by definition is 1/0^3 And you can't divide by 0, Hence, -3 is not a solution

[moeberry]

X is NOT equal to -3 Syber, it is undefined, there are only 3 solutions.

[erikroberts]

x = -3 is not a solution. By substituting -3 in for x, we end up with 0 ^ (-3) = 0 ^ (-3). Then, using order of operations (i.e., PEMDAS):

0 ^ (-3) = 0 ^ (-3)
1 / (0 ^ 3) = 1 / (0 ^ 3)
1 /0 = 1 / 0

At this point, we have to stop because we cannot divide by zero. The result is underfined.

[alejandropulidorodriguez]

i loved the "try to do this mentally" comment

[gianpaolosoligo]

x=-3 isn't a solution because it isn't verified in the domain of the base since the base of the exponential has to be grater of zero

[quantumxgaming]

My request plz😊

The next question should be such type of question that will take more than 1 hrs for me plz 😊

Love from India 🇮🇳 ❤

[andy_in_colorado]

I actually never thought about the issue with 0 raised to a negative power. I just figured as long as it wasn't being raised to 0, then the answer was just 0.

A negative base can definitely be raised to 0 or an even power, so x=0 & x=2 are valid solutions. And if we insist on positive bases, then x=4 is the only answer that fits.

[alextang]

(x^2-9)^x = (x+3)^x
case 1, x=0 then
LHS=1=RHS answer
case 2, x<>0
x^2-9=+/-(x+3)
...
x=-3, 2 or 4 answer
Therefore the answer is x={0, -3, 2, 4}😋😋😋😋😋😋

[seanfraser]

0^-3 is, by definition, 1/0^3 = 1/0 which is undefined. So x=-3 is not a solution

[maaarmaaar]

#sugestion:

If: f(5x) – f(4x) = 2x, then f(x) = ?

[victorchoripapa]

I've only got X=0

Edit: I haven't seen X=4 could be also😅

[AnkushSharma-ilxf]

What level of mats is this I am in 11 th standard how to learn this