# Solve $2^x+2^{-x} = 2$

Need to solve:

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

I can't use substitution in this case. Which is the best approach?

Event in this form I do not have any clue:

$$2^x+\frac{1}{2^x} = 2$$

• Why can't you substitute $y=2^x$ and solve the quadratic equation? – David Quinn Jul 11 '16 at 7:28
• There will be only solution x=0 – Aakash Kumar Jul 11 '16 at 7:29
• – lab bhattacharjee Jul 11 '16 at 7:59

Elucidate the problem by using the substitution $u = 2^x$, then you have $$u + \frac{1}{u} = 2$$

Multiply throughout by $u \neq 0$ to get $$u^2 +1 = 2u \iff u^2 - 2u + 1 = 0$$

This is an easy quadratic to solve, you should get $u = 1$ and hence you need only solve $2^x = 1 \iff x = 0$.

• "I can't use substitution in this case." ?? – Yves Daoust Jul 11 '16 at 7:48
• @YvesDaoust presumably the OP was under the impression that a substitution wouldn't work here. – Zain Patel Jul 11 '16 at 7:49
• yes, my fault I din't think about this case – J.Doe Jul 11 '16 at 8:11

Hint: By AM-GM inequality , $2^x+2^{-x} \geq 2 \times \sqrt{2^x \times 2 ^{-x}} = 2$ When $2 ^x = 2^{-x}$ the equality holds.

Alternative solution:

Easy to show that $x = 0$, is one possible solution.

$$\frac{d(2^x+2^{-x})}{dx} = ( 2^x -2^{-x}) \log 2$$ which positive over $(0,+\infty)$, negative over $(-\infty,0)$

• What a nice way to do it! :-D – Ant Jul 11 '16 at 7:36
• Just instinct. :-) – Zack Ni Jul 11 '16 at 7:40

A contrived solution:

Write

$$\frac{e^{x\ln2}+e^{-x\ln2}}2=1=\cosh(x\ln2),$$

then

$$x\ln2=\cosh^{-1}1=0.$$

Another one:

By inspection, $x=0$ is a solution.

The derivative of the LHS is

$$2^x\ln2-2^{-x}\ln2,$$ which is positive for $x>0$ and negative for $x<0$, so the function is monotonic on both sides and there are no other roots.

Yet another:

Observe

$$2^x-2+2^{-x}=\sqrt2^{2x}-2\sqrt2^x\sqrt2^{-x}+\sqrt2^{-2x}=(\sqrt2^x-\sqrt2^{-x})^2.$$

The only root is when

$$\sqrt2^x=\sqrt2^{-x},$$ i.e. $$x=-x.$$

A last one:

The equation can be rewritten

$$2^x-1=1-2^{-x},$$ i.e. unless the RHS is zero

$$\frac{2^x-1}{1-2^{-x}}=2^x=1.$$

There is no other solution than $x=0$.

Substitute $y=2^x$. Hence, the equation is: $y+y^{-1}=2$ which is equivalent to: $y^2-2y+1=0$. Can you take it from here ? Once you find which $y$('s) satisfy the equation, try to find $x$ such that $2^x=y$.

$$2^x =y$$ $$y+ \frac{1}{y} =2$$ Using AM-GM INEQUALITY $$\frac{({y+ \frac{1}{y}})}{2} \ge \sqrt {y.\frac{1}{y}}$$

$$y+ \frac{1}{y}\ge 2$$

Let, $2^x=y$ then the equation becomes

$y+\frac{1}{y}=2\\ \implies y^2+1-2y=0\\ \implies (y-1)^2=0\\ \implies y=1\\ \implies 2^x=1\\ \implies 2^x=2^0\\ \implies x=0$

• Mr. Downvoter! Please be kind enough to comment where it is wrong. If you have time to downvote you also have time to comment! – MonK Aug 4 '16 at 10:17