The video in a comment: 0:01 Q.) 2x = 10, solve for x. 0:07 Q.) x^2 = 10, solve for x. 0:16 Q.) 2^x = 10, solve for x. Solutions: --> 2x = 10, divide by 2 on both sides. Which gives us, =>x = 5. --> x^2 = 10, take the square root on both sides. Which gives us, =>|x| = √10 =>+-x = √10 =>x = +-√10 Thus, x = +3.162 and x = -3.162 (approx) --> 2^x = 10, take log with base as 2 on both sides. Which gives us, =>log_2(2^x) = log_2(10) =>x*log_2(2) = log_2(2×5) We can cancel the log in the LHS, and use the multiplication->addition property of log in the RHS. =>x = log_2(2) + log_2(5) Once again, =>x = 1 + log_2(5) We can go further and change the log to the common log, using the base change property. =>x = 1 + (log5)/(log2) =>x = 1 + 0.699/0.301 (approx) =>x = 1 + 2.322 (approx) Thus, x = 3.322 I was bored..
@FBIchan
2 ай бұрын
keep the second and third solutions in their form shown in the video please, neither are rational
@TheSpoonThatDied
2 ай бұрын
@@FBIchan It's funny
@nabilal.rawahi98472 ай бұрын
You are amazing.. the best teacher I have ever seen.. I admire your elegant style.
@PinkPastelShark2 ай бұрын
1+log_2(5)
@user-fp4vk9wj8g2 ай бұрын
The best teacher that I ever encountered
@ambasing_omaygot2 ай бұрын
Oh wow that was quick
@shadeyfang85642 ай бұрын
So the reason it’s plus or minus in that case is because the prompt is x^2=10. X can be squared in negative or positive to be 10. When is it only + again?
@danielnoble2835Ай бұрын
Why does (.3^2) + .4 = (.3 + .4) ^2? I cannot figure out what this property is that allows this to be true. It only seems to work with numbers between 0 and 1. And only seems to work when raised to the 2nd power.
@Forgottenbreaded2 ай бұрын
Cool
@rishii2296Ай бұрын
You know *log e (natural logarithm)* can be written as *ln* , but don't know *log 2 (binary logarithm)* can be written as *lb* ?
@utkarshgautam7896Ай бұрын
Hello , how to solve the equation (14x)^(2x) =7? I tried myself and searched everywhere but couldn't find the steps
@allozovsky2 ай бұрын
Please solve *√z = −2* over the field of complex numbers *ℂ.*
@diegocabrales
2 ай бұрын
The equation is √z = -2 (1) or equivalently z^(1/2) = -2 (2) Every square root of a complex number has two roots (counted by multiplicity, that is, it could or couldn't be a double root) according to the fundamental theorem of algebra. Now here we're taking only one of its roots and it's real, therefore the omitted root and z are also real. The omitted root is 2. The Equation (2) is so easy to solve, since only takes squaring it: z = 4 (3)
@allozovsky
2 ай бұрын
Strange Steve claims in his every second video that there are no solutions to such equations (neither real nor complex).
@diegocabrales
2 ай бұрын
@@allozovsky He is probably taking the convention that √z is the principal square root of z, which is equivalent to the plus-signed root of the square root of z. Notice I've said plus-signed root instead of positive root since complex numbers are not in general positive nor negative, but they can be plus- or minus-signed. Maybe that's something some complex analysts would do, but in general √z is equivalent to z^(1/2), which by the fundamental theorem of algebra has two roots.
@allozovsky
2 ай бұрын
@@diegocabrales Seems like it might be so, but I think that convention simply makes no sense when solving irrational equations over the field of complex numbers ℂ, since it rarely gives you any solutions at all.
@diegocabrales
2 ай бұрын
@@allozovsky Yes. I have the Spanish book Luis Bernal González and Genaro López Acedo. "Análisis de Variable Compleja" ("Complex Variable Analysis"), second edition, Universidad de Sevilla (Sevilla University) Publisher, 2015, which makes a clear distinction between the n-th root of z, sqrt[n](z), and the principal n-th root of z, sqrt[n]_p(z), at page 83. It defines in particular the principal n-th root of z as sqrt[n]_p(z) = |z|^(1/n)exp[iarg_p(z)/n] where arg_p(z) is the principal argument of z {defined as the unique argument of z lying on the interval (-π, π]}. It also defines the principal logarithm of z and relates the principal n-th root of z with it in the definition (all at page 83), but it's not relevant now. The thing is that there's a clear distinction between functions and principal functions, so we can all agree on what we're doing, unlike what seems is happening between (at least) us and Steve when talking about the square root of z.
@Brid7272 ай бұрын
now do 2↑ ↑ x = 10
@Fire_Axus
Ай бұрын
i think we will need some mythical functions for this...
@user-er3xq3fm8n2 ай бұрын
School 🚫 KZread✅
@CoffebucksАй бұрын
❤❤❤
@Starplat1num2 ай бұрын
Can you create an equation where you can use it as a wifi password?
Пікірлер: 28
The video in a comment: 0:01 Q.) 2x = 10, solve for x. 0:07 Q.) x^2 = 10, solve for x. 0:16 Q.) 2^x = 10, solve for x. Solutions: --> 2x = 10, divide by 2 on both sides. Which gives us, =>x = 5. --> x^2 = 10, take the square root on both sides. Which gives us, =>|x| = √10 =>+-x = √10 =>x = +-√10 Thus, x = +3.162 and x = -3.162 (approx) --> 2^x = 10, take log with base as 2 on both sides. Which gives us, =>log_2(2^x) = log_2(10) =>x*log_2(2) = log_2(2×5) We can cancel the log in the LHS, and use the multiplication->addition property of log in the RHS. =>x = log_2(2) + log_2(5) Once again, =>x = 1 + log_2(5) We can go further and change the log to the common log, using the base change property. =>x = 1 + (log5)/(log2) =>x = 1 + 0.699/0.301 (approx) =>x = 1 + 2.322 (approx) Thus, x = 3.322 I was bored..
@FBIchan
2 ай бұрын
keep the second and third solutions in their form shown in the video please, neither are rational
@TheSpoonThatDied
2 ай бұрын
@@FBIchan It's funny
You are amazing.. the best teacher I have ever seen.. I admire your elegant style.
1+log_2(5)
The best teacher that I ever encountered
Oh wow that was quick
So the reason it’s plus or minus in that case is because the prompt is x^2=10. X can be squared in negative or positive to be 10. When is it only + again?
Why does (.3^2) + .4 = (.3 + .4) ^2? I cannot figure out what this property is that allows this to be true. It only seems to work with numbers between 0 and 1. And only seems to work when raised to the 2nd power.
Cool
You know *log e (natural logarithm)* can be written as *ln* , but don't know *log 2 (binary logarithm)* can be written as *lb* ?
Hello , how to solve the equation (14x)^(2x) =7? I tried myself and searched everywhere but couldn't find the steps
Please solve *√z = −2* over the field of complex numbers *ℂ.*
@diegocabrales
2 ай бұрын
The equation is √z = -2 (1) or equivalently z^(1/2) = -2 (2) Every square root of a complex number has two roots (counted by multiplicity, that is, it could or couldn't be a double root) according to the fundamental theorem of algebra. Now here we're taking only one of its roots and it's real, therefore the omitted root and z are also real. The omitted root is 2. The Equation (2) is so easy to solve, since only takes squaring it: z = 4 (3)
@allozovsky
2 ай бұрын
Strange Steve claims in his every second video that there are no solutions to such equations (neither real nor complex).
@diegocabrales
2 ай бұрын
@@allozovsky He is probably taking the convention that √z is the principal square root of z, which is equivalent to the plus-signed root of the square root of z. Notice I've said plus-signed root instead of positive root since complex numbers are not in general positive nor negative, but they can be plus- or minus-signed. Maybe that's something some complex analysts would do, but in general √z is equivalent to z^(1/2), which by the fundamental theorem of algebra has two roots.
@allozovsky
2 ай бұрын
@@diegocabrales Seems like it might be so, but I think that convention simply makes no sense when solving irrational equations over the field of complex numbers ℂ, since it rarely gives you any solutions at all.
@diegocabrales
2 ай бұрын
@@allozovsky Yes. I have the Spanish book Luis Bernal González and Genaro López Acedo. "Análisis de Variable Compleja" ("Complex Variable Analysis"), second edition, Universidad de Sevilla (Sevilla University) Publisher, 2015, which makes a clear distinction between the n-th root of z, sqrt[n](z), and the principal n-th root of z, sqrt[n]_p(z), at page 83. It defines in particular the principal n-th root of z as sqrt[n]_p(z) = |z|^(1/n)exp[iarg_p(z)/n] where arg_p(z) is the principal argument of z {defined as the unique argument of z lying on the interval (-π, π]}. It also defines the principal logarithm of z and relates the principal n-th root of z with it in the definition (all at page 83), but it's not relevant now. The thing is that there's a clear distinction between functions and principal functions, so we can all agree on what we're doing, unlike what seems is happening between (at least) us and Steve when talking about the square root of z.
now do 2↑ ↑ x = 10
@Fire_Axus
Ай бұрын
i think we will need some mythical functions for this...
School 🚫 KZread✅
❤❤❤
Can you create an equation where you can use it as a wifi password?
2x=x^2=2^x=10 💀
14 seconds anyone?