Solution for this Logarithmic Equation Recently I was going through a problem from the book Problems in Mathematics - *V Govorov & P Dybov* . (x-2)^(log^2(x-2)+log(x-2)^5-12)=10^2log(x-2) I tried solving by first considering log(x−2) as a variable, say t. Then I expressed (x−2) as 10^t . Then after using some properties of log, I reached till here- 10^(t^3+5t^2-12t)=10^2t or 10^(t^3+5t^2-12t-2)=t Now I have no idea how to approach further. The answer in the references says x=3,102,2+10^(−7)

Marisol Rivers 2022-07-21 Answered
Solution for this Logarithmic Equation
Recently I was going through a problem from the book Problems in Mathematics - *V Govorov & P Dybov* .
( x 2 ) log 2 ( x 2 ) + log ( x 2 ) 5 12 = 10 2 log ( x 2 )
I tried solving by first considering log ( x 2 ) as a variable, say t. Then I expressed ( x 2 ) as 10 t . Then after using some properties of log, I reached till here-
10 t 3 + 5 t 2 12 t = 10 2 t
or
10 t 3 + 5 t 2 12 t 2 = t
Now I have no idea how to approach further. The answer in the references says x = 3 , 102 , 2 + 10 7
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Answers (1)

Sandra Randall
Answered 2022-07-22 Author has 17 answers
Equations such as
f ( t ) = 10 t 3 + 5 t 2 12 t 2 t = 0
cannot be solve using analytical methods and numerical methods, such as Newton, should be used.
As you probably notice, you are looking for the intersection of two curves, namely
y 1 ( t ) = 10 t 3 + 5 t 2 12 t 2
y 2 ( t ) = t
If you plot the functions on the same graph, you should notice a clear intersection around t = 1.9. There is also a root close to t = 0 since, around this value, a Taylor expansion gives
y 1 ( t ) = 1 100 3 25 t log ( 10 ) + O ( t 2 )
which has a negative slope while y 2 ( t ) has a positive slope. Using this expansion gives another estimate close to
t = 1 4 ( 25 + 3 log ( 10 ) ) 0.00783509
So, let us define the overall function
f ( t ) = 10 t 3 + 5 t 2 12 t 2 t
and let us try to find its roots starting from a given estimate t 0 . Newton procedure will update this guess accodring to
t n + 1 = t n f ( t n ) f ( t n )
For the first solution, let us start at t 0 = 0; Newton iterates are then : 0.00783509, 0.00801852, 0.0080186 which is the solution for six significant figures.
For the second solution, let us start at t 0 = 1.9; Newton iterates are then : 1.91187, 1.90970, 1.90959 which is again the solution for six significant figures.
Since, from your changes of variable x = 2 + 10 t , the solutions are then x = 3.01864 and x = 83.2064 which are the values given by Tunococ.

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