Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question

put on hold as off-topic by Najib Idrissi, Fly by Night, M Turgeon, Tomás, Brad yesterday

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Najib Idrissi, Fly by Night, M Turgeon, Tomás, Brad
If this question can be reworded to fit the rules in the help center, please edit the question.

1  
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
2  
You have changed the question in the title? (-1) –  TMM Jul 7 '12 at 20:58
4  
($-1$)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 $y\mapsto (3^y+1-4^y)'$.

share|improve this answer

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.

share|improve this answer
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 : $x = 2$

share|improve this answer
8  
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
3  
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

Not the answer you're looking for? Browse other questions tagged or ask your own question.