faster way : let U = 10^x , we have 1^x + 10^x = 100^x rewritten by 1 + U = U^2 . Solution of 1 + U = U^2 is the golden ratio . so the solution is 10^x = golden ration , so for x , the solution is logbase10 (golden ratio ) . i m not sure the complex solutions were asked
The complex solution works out to x=-0.209 + i*pi/ln(10) where ln(10) is present to take you from base e to base 10. I omitted the 2*n*i*pi/ln(10) for simplicity.
You initially said that you are looking for real numbers and used this to justify 1ˣ = 1 and then later in the calculation decided to let the answers be complex so you actually solved 1 + 10ˣ = 100ˣ not 1ˣ + 10ˣ = 100ˣ.
Область определения. Обе функции являются возрастающими, а значит, можно допустить, что 1^x=1. 10^x:=y, тогда y²-y-1=0. D=5, y=(1±√5)/2=10^x. lgx=lg((1±√5)/2)=lg(1±√5)-lg2. Отсюда х=10^(lg(1±√5)-lg2).
@@justjames27 but log base 10 is written as just "lg" (2 letters), and if so 7:45 if we divide by "lg 10", the denominator will be "lg 10", i.e. 1, therefore we could simply omit it, right? :)
@@justjames27 Yes, "log" often denotes base 10, but it also sometimes means leaving the base unspecified. The only warning is that if he ever tries to calculate a numerical approximation of his exact answer, he has to use the same log base at every spot he wrote "log." I feel if he had meant base 10, he would have simplified log 10 to be 1. Also, like you, I have never seen "lg" to mean log base 10.
@RedAngel11993 it is a convention. I dug up a bit more after the above objection. The consensus seems to be different from what we are both saying, more different from my idea anyway. If there is no index and it is not relevant, log without index is a "general" logarithm without base. If it is relevant it is, like you say, base 10. In this case we can assume is not relevant as it is not referring to any "real world" quantity, nor the question asked for anything specific. I agree the OP should have clarified it somehow
For starters I get one real solution: x1= lg(0,5+0,5*Sqrt(5)) If x is real then 1^x = 1. Substitute a = 10^x => a²-a-1 =0 => a1,2 = 0,5+/-0,5*Sqrt(5) a2 is negative, so 10^x would have to be negative which is not possible for a real value of x, which would also contradict our assumption that 1^x = 1 (which is true only for x being a real number). a1 is > 0 so 10^x = a1 has a real solution, namely lg(a1). [lg = log10]. Not being too good with complex numbers I will not try to pry the complex solution out of the problem an will watch the video instead.
faster way : let U = 10^x , we have 1^x + 10^x = 100^x rewritten by 1 + U = U^2 . Solution of 1 + U = U^2 is the golden ratio . so the solution is 10^x = golden ration , so for x , the solution is logbase10 (golden ratio ) . i m not sure the complex solutions were asked
The complex solution works out to x=-0.209 + i*pi/ln(10) where ln(10) is present to take you from base e to base 10. I omitted the 2*n*i*pi/ln(10) for simplicity.
You initially said that you are looking for real numbers and used this to justify 1ˣ = 1 and then later in the calculation decided to let the answers be complex so you actually solved 1 + 10ˣ = 100ˣ not 1ˣ + 10ˣ = 100ˣ.
A question that's so easy it wouldn't even make it to the initial exam to select the first pool of a country's candidates!
Область определения. Обе функции являются возрастающими, а значит, можно допустить, что 1^x=1. 10^x:=y, тогда y²-y-1=0. D=5, y=(1±√5)/2=10^x. lgx=lg((1±√5)/2)=lg(1±√5)-lg2. Отсюда х=10^(lg(1±√5)-lg2).
What was the base of log? ln is log /e, lg is log /10, and log is your case -- what base does it have?
if a base is not specified, it is assumed to be base 10
@@justjames27 but log base 10 is written as just "lg" (2 letters), and if so 7:45 if we divide by "lg 10", the denominator will be "lg 10", i.e. 1, therefore we could simply omit it, right? :)
@ ive never seen lg to abbreviate log base 10 personally, and ive always been taught that just log means log base 10
@ but also i havent watched the video fully so i dont know. some other people are suggesting log base 2 but idk
@@justjames27 Yes, "log" often denotes base 10, but it also sometimes means leaving the base unspecified. The only warning is that if he ever tries to calculate a numerical approximation of his exact answer, he has to use the same log base at every spot he wrote "log." I feel if he had meant base 10, he would have simplified log 10 to be 1. Also, like you, I have never seen "lg" to mean log base 10.
The solution to the problem is not very difficult for those who are interested in logarithms
The presenter forgot that that log10=1 (8::13 instant).
It's a natural log when there is no index...
@@filipposantarelli7787 In the US, natural log is ln. Log without a base specified is 10.
@@filipposantarelli7787 No, it is the decimal logarithm when there is no index...
@RedAngel11993 it is a convention. I dug up a bit more after the above objection. The consensus seems to be different from what we are both saying, more different from my idea anyway.
If there is no index and it is not relevant, log without index is a "general" logarithm without base. If it is relevant it is, like you say, base 10. In this case we can assume is not relevant as it is not referring to any "real world" quantity, nor the question asked for anything specific.
I agree the OP should have clarified it somehow
log(10)=1.
1^(9)+10^(9)=100 1^(3^2) 1(1^1)+5^5(3^2) 2^3^2^3^(1^1) 2^1^1^3 23(x ➖ 3x+2).
lgΦ
just take antilog at 5:55 , it will save so much time
For starters I get one real solution: x1= lg(0,5+0,5*Sqrt(5))
If x is real then 1^x = 1.
Substitute a = 10^x
=> a²-a-1 =0
=> a1,2 = 0,5+/-0,5*Sqrt(5)
a2 is negative, so 10^x would have to be negative which is not possible for a real value of x, which would also contradict our assumption that 1^x = 1 (which is true only for x being a real number).
a1 is > 0 so 10^x = a1 has a real solution, namely lg(a1). [lg = log10].
Not being too good with complex numbers I will not try to pry the complex solution out of the problem an will watch the video instead.
Can’t u just approximate log 2 to 0.3
0:58 aaaaaand I'm lost
Chat got can’t solve this.