MHB Solve Exponential Equation: 2+√2^x + 2-√2^x =4

  • Thread starter Thread starter Chipset3600
  • Start date Start date
  • Tags Tags
    Exponential
AI Thread Summary
To solve the equation (2+√2)^x + (2-√2)^x = 4, the user introduces a substitution, letting C = (2+√2)^x. This leads to the conclusion that 1/C equals 1/(2-√2)^x, establishing a reciprocal relationship between the two terms. The discussion centers on understanding how this transformation simplifies the original equation. The user seeks clarification on the derivation of this reciprocal relationship. The conversation emphasizes the importance of algebraic manipulation in solving exponential equations.
Chipset3600
Messages
79
Reaction score
0
Hi, I'm having problems to solve this equation, pls help me:

$$\left( 2+\sqrt {2}\right) ^{x}+\left( 2-\sqrt {2}\right) ^{x}=4$$
 
Mathematics news on Phys.org
Chipset3600 said:
Hi, I'm having problems to solve this equation, pls help me:

$$\left( 2+\sqrt {2}\right) ^{x}+\left( 2-\sqrt {2}\right) ^{x}=4$$
Put $$C=(2+\sqrt {2})^x$$ Then $$\frac{1}{C}=\frac{1}{(2-\sqrt {2})^x}$$
NOW MY QUESTION IS, HOW DID I GET THIS?

Hint:
$$(2+\sqrt {2})^x (2-\sqrt {2})^x$$

Regards,
$$|\pi\rangle$$
 
Last edited:
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Back
Top