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For what value of $b$ is the line $y=10x$ tangent to the curve $y=e^{bx}$ at some point in the $xy$-plane? The solution is $10/e$.

The question source is GRE 0568: Q23. I would appreciate the fastest way to solve it.

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What have you tried? –  Chris Eagle Oct 10 '12 at 12:40
    
What is a "GRE 0586 Q23 tangent"? –  Henning Makholm Oct 10 '12 at 12:44
    
Is a standardize test, Q23 is the number of the question. –  inquisitor Oct 10 '12 at 12:46
    
@HenningMakholm: I think it is a number for the test or for a problem. –  Babak S. Oct 10 '12 at 12:46
    
I tried to write down the equation of the tangent line to $e^{bx}$ and solve for $b$, (where first I see that they intersect in some point $x_0$) –  inquisitor Oct 10 '12 at 12:47

1 Answer 1

up vote 3 down vote accepted

For the curve $y=e^{bx}$ you have $y\,'=be^{bx}$; the line $y=10x$ has slope $10$, so it can be tangent to the exponential only at a point where $be^{bx}=10$. Of course this point must lie on the exponential and the straight line, so $10x=e^{bx}$. Substitute $be^{bx}$ for $10$ in this last equation to get $bxe^{bx}=e^{bx}$; dividing by $e^{bx}$ shows that $bx=1$, so $e^{bx}=e$, and $b=\dfrac{10}{e^{bx}}=\dfrac{10}e$.

Alternatively, multiply $10x=e^{bx}$ by $b$ to get $10bx=be^{bx}=10$, deduce that $bx=1$, and proceed as above.

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