SALogic, you have found and solved a neat little challenge here . Congratulations. I think that you enjoy the natural logarithms. One day I will be happy with the Lambert w bits also. But it looks complicated with so much algebra.
x=2 First solution that I discovered Checking: LHS = 2^(x-1) = 2^ (2-1) = 2^1 = 2 = x =RHS Next comment mentions another solution......thinking... x= 1 Second solution Checking: LHS = 2^(x-1) = 2 ^ (1-1) = 2^0 = 1 = x = RHS now to see what is happening here , can these be on a graph? * to show the solution values Plotting y = 2^(x-1) a set of points (0 , 0.5) , (0.5 , 1/sqrt(2) ) , (1, 1) * , (1.5 , sqrt(2)) , (2, 2 ) *, (2.5 , sqrt(8) , (3, 4) At x= 1.5 this function2^(x-1) has curved below the line y=x . The function is exponential , smooth and not as steep as y = e^x but is one place to the left of y= 2^x. With a little rearrangement, below: we can look along y =0 for solutions We could see a turning poiint if we looked at dy/dx of y = 2^(x-1) - x might be close to 1.5, - 0.1. If this is interesting it can be chased with a calculator or a chart could be added to a little spreadsheeting.
2^(x-1)=x By observation, Try x=1 2^(1-1)=1 2^0=1, therefore x=1 Try x=2 2^(2-1)=2 2^1=2, therefore x=2. solved in three minutes. If you draw a graph of the function,the solutions are x=1 and x=2.
SALogic, you have found and solved a neat little challenge here . Congratulations.
I think that you enjoy the natural logarithms.
One day I will be happy with the Lambert w bits also. But it looks complicated with so much algebra.
Lambert W Function also know as product log
Thanks for the tips! ❤
x=2 First solution that I discovered
Checking: LHS = 2^(x-1) = 2^ (2-1) = 2^1 = 2 = x =RHS
Next comment mentions another solution......thinking...
x= 1 Second solution
Checking: LHS = 2^(x-1) = 2 ^ (1-1) = 2^0 = 1 = x = RHS
now to see what is happening here , can these be on a graph? * to show the solution values
Plotting y = 2^(x-1) a set of points (0 , 0.5) , (0.5 , 1/sqrt(2) ) , (1, 1) * , (1.5 , sqrt(2)) , (2, 2 ) *, (2.5 , sqrt(8) , (3, 4)
At x= 1.5 this function2^(x-1) has curved below the line y=x . The function is exponential , smooth and not as steep as y = e^x but is one place to the left of y= 2^x.
With a little rearrangement, below: we can look along y =0 for solutions
We could see a turning poiint if we looked at dy/dx of y = 2^(x-1) - x might be close to 1.5, - 0.1. If this is interesting it can be chased with a calculator
or a chart could be added to a little spreadsheeting.
Very nice! ❤
13:08 . . . why 4/4 and not any n/n?
Because 4 = 2² ❤
How did u know about the two solutions
Every equation of this type has two solutions! ❤
2^(x-1)=x
By observation,
Try x=1
2^(1-1)=1
2^0=1, therefore x=1
Try x=2
2^(2-1)=2
2^1=2, therefore x=2. solved in three minutes.
If you draw a graph of the function,the solutions are x=1 and x=2.
2
Very nice! ❤
X=1
x = 2