solving an exponential equation with different bases, a radical power & a quadratic power
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- Опубликовано: 6 авг 2022
- We will solve an exponential equation that has different bases, a radical power & a quadratic power, namely 3^sqrt(x)=6^(x^2). We will be using the change of base formula for logarithm, rules of exponent, factoring techniques, and more. Subscribe to @bprpmathbasics for more fun algebra equations like this!
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The backing up speed in the beginning though lol
Sorry im slow... how did he get 3 to the power of log 6 base 3...
@@blast8012 Whenever you have a log with a base, its value is the solution for x:
a^x = b; log_a(b) = x
So if you take log_3(6) it's the solution for x in this equation:
3^x = 6; x = log_3(6)
So if you wanna get 6, and 6 is equal to 3^x, and x is equal to log_3(6), 6 is equal to 3^(log_3(6)).
Thanks, I'm re-learning algebra. You're lessons are helpful! I needed it
It is very simple
3^(x)½=6^(x²)
We take the logarithm of both sides.
(X½)Log(3)=x²(log(6)
Move to one side. Factoring from x½
X½[log(3)-x^(3/2)log(6)]=0
So x½=0 x=0
And x=(log(3)/(log(6))⅔
You lost me at "take the logarithm of both sides". Sorry for the skepticism, as it looked like a leap of faith from my perspective. How did that maneuver NOT break the equation?
@@alrycaaeveahexendias1236 If you choose the right starting point, the path to the goal will be shortened.
Nice
Ah yes, since 2^6=4^3
we can take the log of both sides giving us log2(64)=log4(64)
I get that you’re joking but it’s a poor joke since it might legitimately reinforce some people with false understanding.
@@alrycaaeveahexendias1236 doing the same thing to both sides of the equation maintains the equality, because (as expressed by the = sign) the values on both sides are necessarily equal
that was great! Really brought up some important issues when solving
I love this trick! NICE!
By taking log on both sides, it can be solved
Nice one!
You could get the log of both sides from the beginning and it would have been simpler.
Where is that other video?
Here's my solution:
3^sqrt(x) = 6^(x²)
sqrt(x)ln(3) = x²ln(6)
x^(3/2) = ln(3)/ln(6) (assuming x =/= 0)
x = [log_6(3)]^(2/3)
I think this is more legitimate than factoring a zero, since that method assumes the function is continous.
dividing by zero generally is a worse method than factorisation because you remove a solution, so you also gotta plug in for x = 0 (when it's that obvious, but in some more complex cases it might not be). Factoring just assures you that it's a solution. Showing continuity isn't that hard here because log_n(x): R+ --> R (for the real-valued logarithm where n is positive) and everything else is a polynomial
As a teacher do you give these type of questions to your students lol ? In my school (EU), We would never ever face such problems that requires you to actually think deeper... Anyway love the channel !
bro just left you on read 😭😭😭😭
#YTAlgebra
I’m from EU too..we have never been given these type of equations😂😂 but these are really easy once you understand the trick tbh
@@marcmarco1305 I mean yeah in general they train us to compute stuff over and over again, especially with logarithms, you legit feels like a bot
1:48 At this point, instead of squaring both sides, can we multiply both sides by sqrt(x)?
Then how to do next step?
that may complicate it cause it gonna be sone sqrt(x) . (x^2 .....)
Yes you can, but it doesn't help much
(log(6)(3))2/3
awesome
Best intro
Question when you changed the base of 6 to 3^(log6/log3) what is that concept called and where can I learn about it?
simply change of base. I don't know where you can learn about it but it is fairly simple.
a = e^(ln a) = e^(ln a ln b / ln b) = e^(ln b))^(ln a / ln b) = b^(ln a / ln b)
therefore
a = b^(ln a / ln b) should work forall non zero a and b, so long as b isn't 1 (that is, for real values)
Property of Equality for Exponential Equations- this property is useful to solve an exponential equation with the same bases. It says when the bases on both sides of an exponential equation are equal, then the exponents must also be equal. You learn this concept on khan Academy or other youtube videos.
the concept is that
log a to the base b = log a/log b and log a to the base b=1/log b to the base a
amazing!!! still too easy haha
No fair making us wait for the other video
Le soluzioni sono x=0,x=(log3)^2/3,con log in base 6
Love frm India 🇮🇳 ❤
Nice videoo
You don't have to check the solutions since both sides of the equation you squared were non-negative.
1dlate
now demonstrate any complex solution!
i lv u 99.9%
2:29 but after dividing by x the solution x=0 is ruled out
we know that x = 0 is a solution so it doesn't matter that we won't get it in the end
He said that
At first second I thought I'm watching a spin-off to Tenet.
I tried to solve it, and I came up with: x = [(log2 3) / (1 + log2 3)]^(1/3). Does it work too?
AHHHH! SHIT! I must've been mistaken; it doesn't work.
@@ProfAmeen08 at least you tried..have you figure it out?
@@marcmarco1305 Yeah, I guess I knew where my mistake was; thx.
The power 1/3 should be 2/3, otherwise it's OK. Note that 1+log₂3=log₂2+log₂3=log₂6, so your fraction = log₂3/log₂6=log3/log6, same as answer in video.
Shouldn't there be 4 solutions? Since we had a quartic equation and then factored it
By a look, we can say that 0 is definitely a solution.
x^3 = something. 3 roots. So, where is 2 complex friends
isnt x just 0
why does he sound so unusually unenergetic????
* multiplies both sides by 0*
*leaves the class*
This a joke
0
Why so complicated? By inspection, x = 0 is an obvious root. Now consider only nonzero roots. Take logs (to any consistent base) on both sides: sqrt x log 3 = x^2 log 6. Since x not equal to zero by assumption, rearrange to x^(3/2) = log3/log6
So x = (log3/log6)^(2/3), which can also be written more compactly (but less usefully) as [log_6 (3)] ^(2/3) - the subscript means log to base 6.
So the two solutions are x = 0 and x = [log_6 (3)] ^(2/3) = (log3/log6)^(2/3)
DON T