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3 years ago

Can someone help me

Mathematics

1 answer:

Yanka [14]3 years ago

7 0

Answer:

Step-by-step explanation:

if right mark brainlyest

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

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

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:

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

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

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

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

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

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

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

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

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

$2^{x} = 2$

$x\cdot \log 2 = \log 2$

$x = 1$

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

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

7 0

3 years ago

The equation y= 7/2 = 1/2 (x-4) is written in point-slope form. What is the y-intercept of the equation?

Ray Of Light [21]

y= 7/2 = 1/2 (x-4)
its written in the question

4 0

4 years ago

Read 2 more answers

What is the value of h in the figure below? In this diagram, ABAD - ACBD.

andreev551 [17]

Answer:

h =12

Step-by-step explanation:

$In\: \triangle ABC, \: \angle ABC = 90\degree \: \&\:\\ BD\perp AC$

Therefore, by geometric mean property:

$BD = \sqrt {AD\times CD} \\h = \sqrt {(AC-CD) \times CD} \\h = \sqrt {(25-16) \times 16} \\h = \sqrt {9 \times 16} \\h = \sqrt {144} \\\huge \red {\boxed {h = 12}} \\$