Tricky Exponential Problem

[varadarajcuram]

My great respects to this teacher. Learning every day.

[.mdareeb]

Also if u were wondering if X=-7, in case of when exponent is even, that would not be true since then x²-16 will be = 49-16=33, which is odd. Therefore x is not equal to -7

[Gmailcommmmmmmmm]

You are real teacher...born to educate

[thatguyslush]

I’m glad these been popping up lately. Keep it going, algorithm!

[guy_with_infinite_power]

Lucky to find this man right here, expecting more calculus questions from him

[Absurdated]

There are two problems with this explanation.
1) Writing (-4)^f(x) may be legal or illegal depending on your definitions. Intuitively, since (-3.999)^(0.008001) is ill-defined, when talking about *real* x (as opposed to integral x) we may prefer not to consider negative exponent bases. If the equation arose from some physical problem, the negative solution probably corresponds to a non-physical situation and should be discarded anyway.
2) More importantly, solving an equation means finding all roots *and* proving we didn't miss any. Simply saying "here are roots x1, x2, x3, put them into the equation and you'll get an equality" is *not* a solution. One way to do this is to recall that whenever the second expression is defined. But notice that now we demand f(x)>0, so we can write a weaker form (i.e. possibly introducing new solutions but not losing any) instead: log_7(|x|)*(x²-16) = x²-16. Or (x-4)(x+4)(log_7(|x|)-1)=0. Tjhis equation has only _four_ solutions, {-7, -4, 4, 7}; then we should substitute them in the original equation and confirm that only three of them are actual roots.

[alenjohn]

I first learnt to solve this kind of exponential equations from GoTutor-Math Solver channel. He solved also a similar problem with four solutions, where the fourth solutions comes from when the exponents are even numbers.

[what]

This is soo useful. My daughter will get this teached soon, so this is a nice refreshment for me so that I can help her study👍

[toysgames]

If i recall correctly, in order to do that you must first check the condition whether we raise into an integer power or a real power

[GREGGRCO]

Rock On !! That should be explained from the very first day, and looked for in all home work problems. It'll become habit.

[aryangupta]

In this manner, x can be substituted with infinite values.
like x = 7, 4, -4, 5, -5, 6, -6, so on. But the actual value of x=7.
🙃🙃🙃

[BRUH-_-_-]

Love your explanation sir, love from India

[sanjarcode]

The trick (systematic way) to solve such problems is to make possibility trees.

To solve this, let's create a possibility tree - one is x = 7 (ok 1 solution) and the other is x ≠ 7, where the only possibility left is that exponent is 0, so x = 4, -4.

x = -4, 4, 7 are the solutions.

[justabunga]

There are 3 solutions total. The way you solve this is to take the log of both sides. This means (x^2-16)ln(x)=(x^2-16)ln(7). Subtract (x^2-16)ln(7) on both sides and factoring out (x^2-16) gives us (x+4)(x-4)ln(x/7)=0 (from factoring and using properties of logarithms), or x=±4 and 7. However, if you try to graph two of the equations simultaneously, x=-4 isn't graphed because it's not in the domain and the base cannot be negative from the equation y=x^(x^2-16). That's why there are only 2 solutions (x=4 and x=7).

[felixkimani]

Yeah I remember doing these maths problems in Highschool 😂. So fun

[RaquelSantos-hjmq]

This is great. I learned all of this at one time but forgot.

[jaythakkar]

Why is x = -4 a solution ? The base should be a positive real number because we define a^x when a > 0

[rengokukyojuro]

I suggest the correct Method. Take log on both sides. Then bring one term on other side with a - ve sign. You now have (x^2-26)(log x - log 7)=0 and thus 3 solutions.

[mayank]

You could also take log then expression would become more simple

[Sara-lkyr]

It means every time there Is the x in the esponent we have to find also the other/s results. 😅
I am italian... We don't study Math well... 🤪
Thanks 😊🙏