solve this equation for x : $27^x - 43^x -9^{(\frac{1}{2}+x)}=0$ how can we solve this equation? I tried to find it graphically but I found a plenty of intersection points with the axis, how can we express these points.
$\begingroup$
$\endgroup$
20
-
$\begingroup$ Is that $4\cdot 3^x$? $\endgroup$– nonuserNov 2, 2019 at 14:49
-
$\begingroup$ no it's forty three $\endgroup$– MarioNov 2, 2019 at 14:51
-
$\begingroup$ Where did you take it from? $\endgroup$– nonuserNov 2, 2019 at 14:51
-
$\begingroup$ from my classroom. $\endgroup$– MarioNov 2, 2019 at 14:52
-
$\begingroup$ What does that mean? Did the teacher write it up or from book... $\endgroup$– nonuserNov 2, 2019 at 14:53
|
Show 15 more comments
3 Answers
$\begingroup$
$\endgroup$
5
Hint: Make a substitution $t=3^x$, then you get:
$$t^3-4t-3t^2=0$$
I suppose you can finish it now...
If that is realy $43$ then just draw the graph of $f(x)= 27^x - 43^x -9^{(\frac{1}{2}+x)}$ say in Geogebra and you will see the result. And it seems it does not have a solution in that case: 
-
$\begingroup$ I find the answer is $x=-\infty$ in the $43$ case $\endgroup$ Nov 2, 2019 at 15:13
-
-
$\begingroup$ i can write the answer as a domain form (x,-infinity) x is the first point that has y = 0 $\endgroup$– MarioNov 2, 2019 at 15:16
-
-
$\begingroup$ "i can write the answer as a domain form (x,-infinity) x is the first point that has y = 0" Why on earth are you assuming there is a point $\alpha$ so that for all $x < \alpha$ then $y= f(x) = 0$. There's utterly no reason to assume that at all. And many have pointed out there are no points $x$ where $f(x) =0$. $\endgroup$ Nov 2, 2019 at 15:27
$\begingroup$
$\endgroup$
Divide through by $9^x$:
$$3^x - \left(\frac{43}{9}\right)^x = 3.$$
For $x>0$, the left side is negative, so it can't equal $3$. For $x<0$, both $3^x$ and $(49/9)^x$ are between $0$ and $1$, so their difference can't be $3$. There is no real solution to your equation.
answered Nov 2, 2019 at 15:14
$\begingroup$
$\endgroup$
1
For $x\to-\infty$, $f(x)=27^x-43^x-9^{\tfrac{1}{2}+x}$ goes to $0$. If you take the derivative of $f(x)$, it is easy to show that it is negative everywhere, hence, $f(x)$ is a decreasing function. This means that $f$ will always be less than zero for $x\in\mathbb{R}$, so there are no real solutions. If you are looking for complex solutions, I suggest you to rewrite $x=a+bi$, use Euler's formula and proceed from there.
answered Nov 2, 2019 at 15:18
-
$\begingroup$ Well, B. Goddard answer is much more elementary. I don't know why you are not accepting his solution? @Mario $\endgroup$– nonuserNov 2, 2019 at 15:23