Solving an exponential equation from Oxford

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#math #algebra #mathbasics

[freaze]

I just looked at the thumbnail and randomly tried to replace x with 0 and with 1 and luckily solving this in like 1 minute (even tho it's not really a good way of justifying saying "it's evident")

[Mr.FelixBlazTube]

Sir, I am you're big fan from India 🇮🇳 and I love ❤ you're basic mathematics videos
Sir, can you plss solve this exponential equation,
6^(x) - 1 = 4^(x) + 2^(x/2) - 3^(2x)
Find the " x " and number of solutions for this equation.
Sir, plsss make a full video on solving this exponential equation




Lot of love 😘 & support 🫂 from me

[xinpingdonohoe]

I rearranged terms and factored it.
4^x (2^x -1)=4(2^x -1)
Then either 4^x=4, so x=1, or 2^x -1=0, so x=0.

[AnAvidAtheist]

0:15
"Try this first."
*"No."*

[kacperolak]

i have turn the equation to be 2^a + 2^b = 2^c + 2^d
and from that we can have a set of logic statements:
a = c and b = d
or
b = c and a = d
wich gives the same solution

[janda]

I managed to factor it to (4^x - 4)(2^x - 1), giving the two real solutions x=0, 1. I realise now that 4^x-4 is a difference of two swuares and can be further factored, but that leads to the non-real solition

[johnnysallow]

I tried solving it by rewriting in terms of powers of 2: 8^x + 4 = 4^x + 2^(x+2)
2^(3x) + 2^2 = 2^2x + 2^(x + 2). When you solve this, you find that every value of X satisfies the equation. What did I do wrong?

[hritamkashyap]

Sir Please explain limitation of descartes rule of sign

[srinivasanperiakaruppan]

You can even use logarithm of base 2 to find the answer.

[sinekavi]

I have a doubt of ((1-x^7)^1/4 - (1-x^4)^1/7) can you please solve this BPRP?

[WanniGames]

another possibility is to realize that t=2^x must be a power of 2, and quickly realise t = 2 is one of the solutions of 2. then do polynomial devision of (x - 2), and find a quadratic that we can solve from that

[misterme-sports]

log₂(-2) has a solution, but only in the complex world:
First, any number can be represented as r*e^(iθ), Where r is the radius from 0|0, i is the imaginary unit and θ is the angle (measured with the positive number line (from 0|0 to ∞|0)).
This is also where the famous equation e^(iπ)=-1 comes from, since the radius on the plane is 1 and the angle 180° (or π).
So -2 represented that way would be 2*e^(iπ) (radius=2, angle=180°=π).
x=log₂2(2*e^(iπ)) | using the log identity log(a*b)=log(a)+log(b), we get
x=log₂(2)+log₂(e^(iπ)) | using another log identity, log(a^b)=b*log(a), and since log₂(2)=1, we get
x=1+πlog₂(e)i as the final complex result :)


@bprpmathbasics I hope I didn't do any mistakes😅

[toddstephen]

Another way to explain why 2^x = -2 is not possible as a real number is because the range of the exponential function is (0, infinity).

[adamoksiuta]

I think t substitution is not necessary. You can factor 2^2x from 2^3x and 2^2x and factor -4 from the rest. You get 2^2x(2^x-1) and -4(2^x-1). Then factor (2^x-1) to get equation (2^2x-4)(2^x-1)=0.

[shdowchasr]

I bet that is a "Make sure you read the question properly" question.

[tglmkyoutube]

Teacher plz teacher

Integrity [sinx+sinx tanx + secx]/[sin^2x tan^2x + 2(1+tanx)] dx

[neilgerace]

B and C are both correct. The equation has one real solution. It also has another real solution.

[MadaraUchihaSecondRikudo]

Intuitively before watching: This is a cubic in terms of 2^x, so expected to be 3 real solutions, by observation 1 and 0 are solutions, my guess is the third would be 2^x = negative which isn't real, so answer is c

[anigami]

i put x=0 and x=1 and they satisfied equation and