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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 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
@Henning: GR0568 is the current version of the practice exam for the Graduate Record Examination in mathematics. If you’re really curious, you can find it here. Most U.S. grad schools require it. – Brian M. Scott Oct 10 '12 at 13:09
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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