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

2 Answers
May 13, 2018

#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.

May 13, 2018

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:

www.desmos.com/calculator

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#.