How do you solve #x+ \log _ { 2} ( x - 7) = 3#?

Question

550 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-01-07T18:04:17

See below.

This type of equations can be solved using the so called Lambert function

#x + Log_2(x - 7) = 3 rArr 2^x (x-7)=2^3# now calling #y = x-7#

#2^(y+7)y=2^3# or

#2^y y = 2^3/2^7 = 2^-4#

but #2^y = e^(lambda y)# then

#e^(lambda y)(lambda y)/lambda = 2^-4# or

#e^(lambda y)(lambda y) = 2^-4lambda#

Now #Y = X e^X rArr X = W(X)# then

#lambda y = W(2^-4 lambda) rArr x = (W(2^-4 lambda))/lambda +7#

Here #lambda = log_2 2#

and finally

#x = 7.05995584253551#