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The number of real solutions of the equation $e^x=3x$ is ______.

I have checked the solution in desmos.com the answer is "2".

But I am not able to solve it

I tried using $x=\ln(3x)$ but could not proceed further.

As there any general method I could find the number of intersection of the equation $e^x=ax$ where $a>0$

Math Attack
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Samar Imam Zaidi
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    Compute the derivative, observe that it has a unique zero, and then that the function is negative at that point. This implies it has exactly two zeros. – Stephen Nov 12 '23 at 11:46
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    This is one of those cases where you can't feasibly calculate the solutions. You have to argue about their number without knowing their precise whereabouts. Use the intermediate value theorem combined with the behavior of $e^x-3x$ at the infinities and at its minimum. – Vercassivelaunos Nov 12 '23 at 11:55

2 Answers2

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The derivative of $f(x)=e^x-ax$ is $e^x-a$. So the function has an extremal point at $x=\ln(a)$. the derivative is positive for $x>\ln(a)$ and negative for $x<\ln(a)$, so $x=\ln(a)$ is a minimum. The function is a sum of convex functions, so it is convex, i.e. the minimum is global

If $a>e$ $f(\ln(a))$ is negative, so there are exactly 2 intersections.

If $a=e$ $f(\ln(a))=0$ so it is the only intersection

If $a<e$ $f(\ln(a))$ is positive, so there are no intersections

Marco
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Let, $y=e^x$ and $y=px$ curves touch at one point $A(\alpha,\beta)$ (Where, p is a const.)

Equation of tangent on $e^x$ at the point $(\alpha,\beta)$,

Now, $(\alpha,\beta)$ lies on $y=e^x$ curve, $\beta=e^\alpha$

Slope of the tangent is $e^\alpha$.

Now, $e^\alpha=\frac{\beta}{\alpha}\Rightarrow \alpha=1\quad(\because \beta=e^\alpha)$

So, $p=e^1=e$

enter image description here.

for, $p>e$ two solutions solution

$p=e$ exact one solution

$0<p<e$ no solutions.

O M
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