# solve this equation for x : $27^x - 43^x -9^{(\frac{1}{2}+x)}=0$

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.

• Is that $4\cdot 3^x$? Nov 2, 2019 at 14:49
• no it's forty three Nov 2, 2019 at 14:51
• Where did you take it from? Nov 2, 2019 at 14:51
• from my classroom. Nov 2, 2019 at 14:52
• What does that mean? Did the teacher write it up or from book... Nov 2, 2019 at 14:53

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:

• I find the answer is $x=-\infty$ in the $43$ case Nov 2, 2019 at 15:13
• I did, I found this : ibb.co/fqXZ5hH Nov 2, 2019 at 15:13
• i can write the answer as a domain form (x,-infinity) x is the first point that has y = 0 Nov 2, 2019 at 15:16
• And i'm searching about the first point Nov 2, 2019 at 15:16
• "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 $f(x) =0$. Nov 2, 2019 at 15:27

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.

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.

• Well, B. Goddard answer is much more elementary. I don't know why you are not accepting his solution? @Mario Nov 2, 2019 at 15:23