Is Figure 1 similar to Figure 2?

Mathematics

1 answer:

tamaranim1 [39]4 years ago

6 0

Answer:

no figure 1 isn't similar to figure 2

Step-by-step explanation:

because its sides aren't the same numbers nor do the numbers correlate with the numbers in figure one its 2 different numbers

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

fomenos

The <em>polynomial-like</em> expression is satisfied by the <em>real</em> value <em>x = 1</em>.

<h3>How to determine the real solution of a polynomial-like expression</h3>

In this question we must apply the concepts of logarithms and <em>algebra</em> properties to solve the <em>entire</em> expression. Initially, we expand the right part of the expression:

$(2^{x}-4)^{3}+(4^{x}-2)^{3} = (4^{x}+2^{x}-6)^{3}$

$(2^{x}-4)^{3} + (4^{x}-2)^{3} = [(2^{x}-4)+(4^{x}-2)]^{3}$

$(2^{x}-4)^{3}+(4^{x}-2)^{3} = (2^{x}-4)^{3}+3\cdot (2^{x}-4)^{2}\cdot (4^{x}-2)+3\cdot (2^{x}-4)\cdot (4^{x}-2)^{2}+(4^{x}-2)^{3}$

$3\cdot (2^{x}-4)^{2}\cdot (4^{x}-2)+3\cdot (2^{x}-4)\cdot (4^{x}-2)^{2} = 0$

$2^{x}-4 + 4^{x}-2 = 0$

$2^{x}\cdot 2^{x} + 2^{x}-6 = 0$

$u^{2}+u - 6 = 0$

$(u+3)\cdot (u-2) = 0$

Hence, the roots of the pseudopolynomial are $u_{1} = -3$ and $u_{2} = 2$ . Only the second one have a real value of <em>x</em>. Hence, we have the following solution: