1
$\begingroup$

Here's the formula: $$ x^{\ln \left( 3 \right)}-3^{\ln \left( x \right)}$$

I know it's equal to $0$ because I've tried different values for $x$, but how do I solve it, how do I simplify it to $0$?

$\endgroup$
1

4 Answers 4

7
$\begingroup$

First $$ x^{\ln(3)}=e^{\ln(3)\ln(x)} $$ And $$ 3^{\ln(x)}=e^{\ln(x)\ln(3)} $$

So yes it values $0$.

$\endgroup$
2
  • $\begingroup$ it's funny that I can solve harder formulas but the most simple I can not. Any tips on how to overcome that ? Practice or memorize? ;) $\endgroup$
    – n00b
    Mar 4, 2018 at 0:20
  • 2
    $\begingroup$ Practice is better IMO. $\endgroup$ Mar 4, 2018 at 0:33
6
$\begingroup$

Hint: For any positive real number $r$, $r=e^{\ln (r)}$. Apply this now to $r=x$ and $r=3$.

$\endgroup$
5
  • 1
    $\begingroup$ Ah, I see three of us have posted almost simultaneously! $\endgroup$ Mar 4, 2018 at 0:00
  • $\begingroup$ What is the problem? $\endgroup$
    – n00b
    Mar 4, 2018 at 0:14
  • $\begingroup$ @user32073, I'm afraid I don't understand your comment. $\endgroup$ Mar 4, 2018 at 0:18
  • $\begingroup$ what is the problem about the posts being posted almost simultaneously? $\endgroup$
    – n00b
    Mar 4, 2018 at 0:19
  • $\begingroup$ @user32073, oh, there's no problem, I was just acknowledging that three of us had almost the exact same idea -- a case of great minds thinking alike.... $\endgroup$ Mar 4, 2018 at 0:23
3
$\begingroup$

Hint: use the fact that $a^x=e^{x\ln a}$ for $a>0$.

$\endgroup$
2
$\begingroup$

Note that for positive values of $x$,$$ x^{\ln \left( 3 \right)}=3^{\ln \left( x \right)}= e^{ln(3).ln(x)}$$

Therefore, $$ x^{\ln \left( 3 \right)}-3^{\ln \left( x \right)}=0$$

$\endgroup$
4
  • 2
    $\begingroup$ You are initially assuming the answer equals $0$. Why? $\endgroup$
    – user535339
    Mar 4, 2018 at 0:01
  • $\begingroup$ @idk Because making an assumption is a good first step of proving statements (especially by contra-diction). For example, we may not know if $\sqrt 2$ is rational, or irrational. So, we assume $\sqrt 2 = p/q$ and where does this lead us? $\endgroup$
    – Mr Pie
    Mar 4, 2018 at 0:07
  • 1
    $\begingroup$ @user477343 What you are believing is incorrect. Here, you have to prove $a-b=0$, and the answer is assuming $a-b=0$ to prove $a-b=0$. In the case of $\sqrt 2$, you are assuming the OPPOSITE of what is correct to prove what is correct. $\endgroup$
    – user535339
    Mar 4, 2018 at 0:30
  • $\begingroup$ @idk ahhh I see... thank you for letting me know :) $\endgroup$
    – Mr Pie
    Mar 4, 2018 at 1:09

Not the answer you're looking for? Browse other questions tagged .