The polynomial-like expression is satisfied by the real value x = 1.
In this question we must apply the concepts of logarithms and algebra properties to solve the entire expression. Initially, we expand the right part of the expression:
[tex](2^{x}-4)^{3}+(4^{x}-2)^{3} = (4^{x}+2^{x}-6)^{3}[/tex]
[tex](2^{x}-4)^{3} + (4^{x}-2)^{3} = [(2^{x}-4)+(4^{x}-2)]^{3}[/tex]
[tex](2^{x}-4)^{3}+(4^{x}-2)^{3} = (2^{x}-4)^{3}+3\cdot (2^{x}-4)^{2}\cdot (4^{x}-2)+3\cdot (2^{x}-4)\cdot (4^{x}-2)^{2}+(4^{x}-2)^{3}[/tex]
[tex]3\cdot (2^{x}-4)^{2}\cdot (4^{x}-2)+3\cdot (2^{x}-4)\cdot (4^{x}-2)^{2} = 0[/tex]
[tex]2^{x}-4 + 4^{x}-2 = 0[/tex]
[tex]2^{x}\cdot 2^{x} + 2^{x}-6 = 0[/tex]
[tex]u^{2}+u - 6 = 0[/tex]
[tex](u+3)\cdot (u-2) = 0[/tex]
Hence, the roots of the pseudopolynomial are [tex]u_{1} = -3[/tex] and [tex]u_{2} = 2[/tex]. Only the second one have a real value of x. Hence, we have the following solution:
[tex]2^{x} = 2[/tex]
[tex]x\cdot \log 2 = \log 2[/tex]
[tex]x = 1[/tex]
The polynomial-like expression is satisfied by the real value x = 1. [tex]\blacksquare[/tex]
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