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Please help me for solution of this equation: $$(\sqrt 3)^x + 1 - 2^x = 0$$

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What type of solution are you looking for? There are no real solutions, since $\sqrt{2}^x < \sqrt{3}^x + 1$ for all $x$. –  Cocopuffs Jul 7 '12 at 20:52
You have changed the question in the title? (-1) –  TMM Jul 7 '12 at 20:58
($-1$)What kind of game is this? Commenting on other answers that a special answer is correct (not only once!), never heard of it. –  Gigili Jul 7 '12 at 21:14

3 Answers 3

There is no real solution because $3^y+1>2^y$ for all real values of $y$ ($y=\frac x2$). For $y$ positive this is true because $3^y+1>3^y>2^y$, for $y$ negative because of the $+1$.

Numerical complex solutions may be found for $y$.

For the modified problem (studying $(\sqrt 3)^x + 1 - 2^x = 0$) use the same variable $y=\frac x2$ getting : $$3^y+1=4^y$$
This has only one (trivial) solution $y=1$ (corresponding to $x=2$) as may be seen by studying the function derivative $y\mapsto (3^y+1-4^y)'$.

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Note: The solution below deals with the original question.

If $a\gt b\gt1$, then the equation $a^x+1-b^x=0$ has no real solution $x$. (Your question is the case $a=\sqrt3$, $b=\sqrt2$.)

Here is a proof. First, if $x\geqslant0$, then $a^x\geqslant b^x$ hence $a^x+1-b^x\geqslant1$ and $x$ is not a solution. Second, if $x\lt0$, then $b^x\lt1\lt1+a^x$ hence $x$ is not a solution either.

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up vote -5 down vote accepted

The given equation is:

$(\sqrt 3)^x + 1 - 2^x = 0$

Given equation can remark the like:

$(\sqrt 3)^x + 1 = 2^x$.

Divide both side by $2^x$, by means:

$(\sqrt 3)^x + 1 = 2^x /: 2^x$

$\frac{(\sqrt 3)^x}{2^x} + \frac{1}{2^x}=\frac{2^x}{2^x}$

$(\frac{1}{2})^x + (\frac{\sqrt{3}}{2})^x$ = 1.

Since $\sin 30^0 = \frac{1}{2}$ and $\cos 30^0 = \frac{\sqrt{3}}{2}$, have:

$\sin^x 30^0 + \cos^x 30^0 = 1$.

Taking into account the basics trigonometry identity ($\sin^2 \alpha + \cos^2 \alpha = 1$) have : $x = 2$

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This is incorrect. $$(\sqrt{3})^2+1-(\sqrt{2})^2=3+1-2=2\neq0$$ Your error was that $$\frac{(\sqrt{3})^x}{(\sqrt{2})^x}=\left(\frac{\sqrt{3}}{\sqrt{2}}\right)^x\neq‌​\left(\frac{\sqrt{3}}{2}\right)^x.$$ –  Zev Chonoles Jul 7 '12 at 20:51
(And a quick sanity check would have told you that $x = 2$ leads to $3 + 1 - 2 \neq 0$...) –  TMM Jul 7 '12 at 20:57
No, this is correct solution –  Madrit Zhaku Jul 7 '12 at 21:06
It is a quite incomplete (but fixable) solution of the revised problem, not of the original problem. –  André Nicolas Jul 7 '12 at 21:12

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