Solving the equation $2^x+2^{-x} = 10$

Question

I'm trying to figure out how to solve this equation in terms of $x$.

Mistake I made so far is:

$$\log(2^x) + \log(2^{-x}) = \log(10)$$

$$x \cdot \log(2) - x\cdot \log(2) = 1$$

$$0 = 1$$

Answer 1

To make this clearer, make the substitution $u = 2^x$. Then your equation becomes

$$u + \frac 1 u = 10$$

Multiply through by $u$:

$$u^2 + 1 = 10u$$

Bring everything to one side:

$$u^2 - 10u + 1 = 0$$

You can use the quadratic formula to find roots. Keep in mind that $u = \text{stuff}$ is what you're finding. Thus, since $x = \log_2 (u)$, you will then take the $\log_2$ of the solutions of the quadratic to find $x$.

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Also, to address your mistake:

$$\log_b(x+y) \neq \log_b(x) + \log_b(y)$$

When you took the log of both sides, you cannot apply it term-by-term, you have to apply it to the entire side of the equation. The property you might be thinking of is

$$\log_b(xy) = \log_b(x) + \log_b(y)$$

Answer 2

Let $2^x=y$

The equation becomes: $y^2-10y+1=0$. This quadratic equation has two solutions:

$$y = \frac{10 \pm \sqrt{100-4}}{2} = 5 \pm 2\sqrt{6}.$$

So you get $x = \log_2(5+2\sqrt{6})$ and $x = \log_2(5-2\sqrt{6})$ as solutions.

Answer 3

Welcome to math.stackexchange!

Hint: $$2^x + 2^{-x} = 10$$ is equivalent to: $$2^{x} + \dfrac{1}{2^{x}} = 10$$ and we can clear the fraction by multiplying on both sides by $2^{x}$.