How do you solve #\log _ { 2} ( x ^ { 3} + x ^ { 2} + 1) = 6#?

1 Answer
George C.
Mar 31, 2018

Real root:

#x = 1/3(-1+root(3)((1699+9sqrt(35637))/2)+root(3)((1699-9sqrt(35637))/2))#

and related complex roots.

Explanation:

Given:

#log_2(x^3+x^2+1) = 6#

Taking exponents base #2# of both sides, we get:

#x^3+x^2+1 = 2^6 = 64#

Subtract #64# from both sides to get a cubic equation in standard form:

#x^3+x^2-63 = 0#

We can solve this cubic using standard methods...

Given:

#f(x) = x^3+x^2-63#

Discriminant

The discriminant #Delta# of a cubic polynomial in the form #ax^3+bx^2+cx+d# is given by the formula:

#Delta = b^2c^2-4ac^3-4b^3d-27a^2d^2+18abcd#

In our example, #a=1#, #b=1#, #c=0# and #d=-63#, so we find:

#Delta = 0+0+252-107163+0 = -106911#

Since #Delta < 0# this cubic has #1# Real zero and #2# non-Real Complex zeros, which are Complex conjugates of one another.

Tschirnhaus transformation

To make the task of solving the cubic simpler, we make the cubic simpler using a linear substitution known as a Tschirnhaus transformation.

#0=27f(x)=27x^3+27x^2-1701#

#=(3x+1)^3-3(3x+1)-1699#

#=t^3-3t-1699#

where #t=(3x+1)#

Cardano's method

We want to solve:

#t^3-3t-1699=0#

Let #t=u+v#.

Then:

#u^3+v^3+3(uv-1)(u+v)-1699=0#

Add the constraint #v=1/u# to eliminate the #(u+v)# term and get:

#u^3+1/u^3-1699=0#

Multiply through by #u^3# and rearrange slightly to get:

#(u^3)^2-1699(u^3)+1=0#

Use the quadratic formula to find:

#u^3=(1699+-sqrt((-1699)^2-4(1)(1)))/(2*1)#

#=(1699+-sqrt(2886601-4))/2#

#=(1699+-sqrt(2886597))/2#

#=(1699+-9sqrt(35637))/2#

Since this is Real and the derivation is symmetric in #u# and #v#, we can use one of these roots for #u^3# and the other for #v^3# to find the Real root:

#t_1=root(3)((1699+9sqrt(35637))/2)+root(3)((1699-9sqrt(35637))/2)#

and related Complex roots:

#t_2=omega root(3)((1699+9sqrt(35637))/2)+omega^2 root(3)((1699-9sqrt(35637))/2)#

#t_3=omega^2 root(3)((1699+9sqrt(35637))/2)+omega root(3)((1699-9sqrt(35637))/2)#

where #omega=-1/2+sqrt(3)/2i# is the primitive Complex cube root of #1#.

Now #x=1/3(-1+t)#. So the roots of our original cubic are:

#x_1 = 1/3(-1+root(3)((1699+9sqrt(35637))/2)+root(3)((1699-9sqrt(35637))/2))#

#x_2 = 1/3(-1+omega root(3)((1699+9sqrt(35637))/2)+omega^2 root(3)((1699-9sqrt(35637))/2))#

#x_3 = 1/3(-1+omega^2 root(3)((1699+9sqrt(35637))/2)+omega root(3)((1699-9sqrt(35637))/2))#

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