Question

How do you solve for x in `x+log_2x=6` ?

Answer 1

`x=4`

Explanation:

This is a transcendental equation since x is in the logarithm and not. Therefore, it does not in general have an easy solution and there isn't any perfect method for solving it.

This one does have a simple solution, but this is because this function had to be selected very carefully.

Let's assume there's some integer `n` such that `2^n = x`. The equation then becomes
`2^n + n = 6`

Plugging in values of n,
`n = 0 implies 2^n + n = 1 ne 6`
`n = 1 implies 2^n + n = 2 ne 6`
`n = 2 implies 2^n + n = 6 = 6`

So the solution `x=4` happens to be the right solution.

Answer 2

One cannot solve the equation `x+log_2x=6` using algebraic methods.

Explanation:

One can use a graphical method by converting the base to base `e` and then separating into two equations.

Convert to base `e`:

`x+ln(x)/ln(2)=6`

Subtract x from both sides:

`ln(x)/ln(2)=6-x`

Multiply both sides by `ln(2)`

`ln(x)=6ln(2)-xln(2)`

Separate into two equations:

`color(red)(y = ln(x))` and `color(blue)(y= 6ln(2)-xln(2))`

Graph the two equations:

Please observe that the two graphs intersect at `x = 4`

You can use a recursive computation method such as Newton's Method to approximate the solution. Because the method requires many lines of computation, I will not do it, here, but I have provided a link so that you may read about it.

The easiest way to obtain a solution is to enter the original equation into WolframAlpha

Please open the above link and observe that WolframAlpha has computed the solution and it is `x = 4`.