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Find all real $x$ that satisfy $(2^x - 4)^3 (4^x - 2)^3 = (4^x 2^x - 6)^3. $

Sagot :

xero099

The polynomial-like expression is satisfied by the real value x = 1.

How to determine the real solution of a polynomial-like expression

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]

To learn more on logarithms, we kindly invite to check this verified question: https://brainly.com/question/24211708

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