Help with Summer Calc HW?

Question

I have done everything I could think of to this problem, but everything I do either makes the problem undefined or equal to zero. Anyone help?

(2^x) + (2^-x) = 5

Solve for x.

Answer 1

First, make the equation look easier:

Let y = 2^x

Since (2^-x) = 1/(2^x), we can put y in for 2^x as follows:

y + 1/y = 5

Multiply both sides by y:

y^2 + 1 = 5y

Rearrange into a quadratic equation:

y^2 - 5y + 1 = 0

Solve this as a normal quadratic equation and find the roots:

y =  4.7913 or y =  0.20871

Remember that y = 2^x

2^x = 4.7913 or 2^x = 0.20871

Natural log of both sides of each result:

ln(2^x) = ln(4.7913) and ln(2^x) = ln(0.20871)

Rearrange the ln() using its power properties:

x(ln(2)) = ln(4.7913) and x(ln(2)) = ln(0.20871)

Rearrange:

x = ln(4.7913)/ln(2) and x = ln(0.20871)/ln(2)

Calculator work on ln(2) gives you:

x = 2.26 and x = -2.26

I think that will do it...the key is to reduce complexity.  You should feel free to use substitutions of a single variable for a function to clarify the equation, in this case into a simple quadratic equation.  Also, remember the tool of taking the ln of both sides.  That's especially useful when dealing with powers of variables, because the algebraic properties of natural logs are very exponent friendly.

Good luck!