What I like about this is that it's easy to see the x=2 solution, but work is required to extract the other solution. I always try and work these out in my head, but this was too much 😅. Thanks!!!
You skipped the naive method, which in this case was probably best. 100 is 5^2*2^2 and one of the multiplicants on the left is 5^x so try x=2. That works. So then do the log to find the other solutions and realize that x=2 has to be one of the factors and synthetic division will give you the other.
What I like about this is that it's easy to see the x=2 solution, but work is required to extract the other solution.
I always try and work these out in my head, but this was too much 😅. Thanks!!!
Hearing this "Hello, everyone" makes my mind shift into math mode. I'm being Pavlov trained.
Very nice!!!! I liked it! I used the first method to solve it
You skipped the naive method, which in this case was probably best. 100 is 5^2*2^2 and one of the multiplicants on the left is 5^x so try x=2. That works. So then do the log to find the other solutions and realize that x=2 has to be one of the factors and synthetic division will give you the other.
Have I had enough coffee this morning?
Let's watch a SyberMath video and find out.
We've done lots of those. Can we do loop integrals, partial differential equations, quaternions, n-dimensional manifolds, or boolean logic now?
2 ^ (3x /(x+1) -2) times 5 ^(x-2) = 1
This consists of two different equations
3x/(x+1) = 2
And x = 2
Both of them are consistent and leads to x= 2
Fantastic question...👌👌👌.
9:27 reverse cross multiplication. I guess we can call it "uncross division"?
very short solution: 8^(x/x+1).5^x= 25*4= (5^2)(2^2)
2^3(x/x+1).5^x= (5^2)(2^2)
so x=2 and 2^3(x/x+1)= 2 hence x=2
There is an easy way of solving these in general by doing e^(ln(8)*x/(x+1)+ln(5)x)=e^ln(100) and then the rest is very obvious.
Nice!
x=2 or x=-1/log5
x=2 is the first solution. I can use log to get an 'ugly' solution
4*25=100
x = 2
👏
2
x=2, -1/log5
-1 is invalid cause you get -1/0
@@yuvanmar42 what??
X=2 but nice problem
But that's only one solution. Do the math to obtain the other one.
Is that your final answer?
@@mcwulf25 no no there are two solution. I worked out the 1st answer mentally.