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:

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


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 $2^x = 1 \iff x = 0$.

  • $\begingroup$ "I can't use substitution in this case." ?? $\endgroup$ – Yves Daoust Jul 11 '16 at 7:48
  • $\begingroup$ @YvesDaoust presumably the OP was under the impression that a substitution wouldn't work here. $\endgroup$ – Zain Patel Jul 11 '16 at 7:49
  • $\begingroup$ yes, my fault I din't think about this case $\endgroup$ – 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 $(-\infty,0)$

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

A contrived solution:





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:



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


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

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

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

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