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

Question

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.

Answer 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.

Answer 2

Hint: Make a substitution $t=3^x$, then you get:

$$t^3-4t-3t^2=0$$

I suppose you can finish it now...

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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:

Answer 3

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.