Question

How do you solve `7^(x - 2) = 5x`?

Answer 1

#0.0041144, nearly.

Explanation:

The difference between the RHS and LHS changes sign in (0. .5).

0 appears to be closer to the zero of the equation.

Rearrange to the form `x=7^(x-2)/5=7^x/(7^2X5)=7^x/ 245` that

befits application of an iterative method, for successive

approximations, with a starter guess-value

Now, use the discrete analogue

`x_n=7^(x_(n-1))/245, n= 1, 2, 3, ..`, with starter `x_0=0`.

This is of the form `x_n=f(x_(n-1)` and is an example for a non-linear

difference equation.

The sequence of approximations is

0.00408... 0.00414... 0.004114 0.0041144..., for n = 1, 2, 3, 4, ., with

the starter `x_0=0`..

For x = 0.004144, `x-7^x/245=O(10^(-8))`.

Despite that we get a good approximation to the solution of the given equation, the sequence tends to the solution of the discrete analogue and not the solution of the given equation, in mathematical exactitude. Yet, we get closer, when we advance, and the difference is bounded, with a limit that is numerically `>=0`.
This analysis is exclusive, for every such discrete analogue, in respect of every such equation..

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