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.
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Is that $4\cdot 3^x$? – nonuser Nov 02 '19 at 14:49
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no it's forty three – Mario Nov 02 '19 at 14:51
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Where did you take it from? – nonuser Nov 02 '19 at 14:51
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from my classroom. – Mario Nov 02 '19 at 14:52
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What does that mean? Did the teacher write it up or from book... – nonuser Nov 02 '19 at 14:53
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the teacher write it up. – Mario Nov 02 '19 at 14:54
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Ask your friends if they also have 43 – nonuser Nov 02 '19 at 14:54
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Yes I asked them :) – Mario Nov 02 '19 at 14:55
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Ok, then ask the teacher. You do have his/her email? – nonuser Nov 02 '19 at 14:56
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unfortunately, I haven't his email. – Mario Nov 02 '19 at 14:57
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How do you know he didn't write $4\cdot 3^x$? If I write this down you can hardly say if it is $43^x$ or $4\cdot 3^x$. – nonuser Nov 02 '19 at 14:58
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I don't understand you – Mario Nov 02 '19 at 15:00
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Can you try solving it? – Mario Nov 02 '19 at 15:02
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@Mario Aqua is saying that you may be misinterpreting the problem. If the middle term is $4\cdot 3^x$, the problem has a "relatively nice" result. If the middle term is $43^x$, the problem does not have a "relatively nice" result. So, if you have only been looking at problems with "nice" results, it is more likely that you have a typo, and the problem is actually what Aqua suggests. – SlipEternal Nov 02 '19 at 15:03
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I did, can you answer the previous question please? – nonuser Nov 02 '19 at 15:03
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Ok can we suppose that the middle term is $43^x$ – Mario Nov 02 '19 at 15:04
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what will be the first value of x at which the equation will equal to zero – Mario Nov 02 '19 at 15:05
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@Aqua which question that you want me to answer it – Mario Nov 02 '19 at 15:08
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If the middle term is $43^x$, then it has no real solutions: Wolframalpha shows no real solutions Do you want Complex solutions? – SlipEternal Nov 02 '19 at 15:11
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The probability that the middle term is $4\cdot3^x$ is $100%$. Listen to InterstellarProbe and Aqua. – Nov 02 '19 at 15:18
3 Answers
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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: 
nonuser
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i can write the answer as a domain form (x,-infinity) x is the first point that has y = 0 – Mario Nov 02 '19 at 15:16
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"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 $f(x) =0$. – fleablood Nov 02 '19 at 15:27
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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.
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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.
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Well, B. Goddard answer is much more elementary. I don't know why you are not accepting his solution? @Mario – nonuser Nov 02 '19 at 15:23