# Cambrige Entry Examination

I recently came across a video with the problem to find the sum of the solutions of

$$3^x – \sqrt{3}^{(x+4)} + 20 = 0$$

Maybe anybody who is intimate with this kind of computations sees the trick. You can set

$$y = \sqrt{3}^x,$$

and get the equivalent equation

$$y^2-9y+20=0.$$

Solving this, you get

$$y = \sqrt{3}^x = 3^{x/2} = \dfrac{9 \pm 1}{2}.$$

Thus

$$x_1+x_2 = 2 \log_3(20) = \dfrac{2 \ln(20)}{\ln(3)}.$$

That’s not very exciting.

But why are they asking to find the sum of the solutions? If you hear that, you might think of the Vieta’s fact that the sum and the product of the solutions of a quadratic equation are the coefficients. Using this, you get

$$y_1y_2 = \sqrt{3}^{(x_1+x_2)} = 20$$

immediately.

That’s were I decided to write about this problem here. E.g., you might change the original problem to

$$3^x – \sqrt{3}^{(x+4)} + 9 = 0$$

and get the much more pleasing solution

$$x_1 + x_2 = 2 \log_3(9) = 4.$$

Nice trick! But it is dangerous. Applying the trick to

$$3^x – \sqrt{3}^{(x+4)} + 81 = 0$$

yields the false sum 8.

In fact, this equation does not have a solution at all! Unless you allow complex numbers, and are careful with complex logarithms. In that case, you get multiple solutions.

So be careful with tricks!

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