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Solve the following equation. $$3^{x^2-6x+11}=8+\cos^2\frac{\pi x}{3}.$$

enter image description here Here is the problem that I am struggling with it I tried to take a logarithm of both side but I kindda stuck can someone help me with this? Thanks

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$$9\geq8+\cos^2\frac{\pi x}{3}=3^{x^2-6x+11}=3^{(x-3)^2+2}\geq9.$$ Thus, $\cos^2\frac{\pi x}{3}=1$ and $x=3$, which gives the answer: $$\{3\}$$

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Add 1 to each side of the equation to get

$$3^{x^2-6x+11} + 1 = 9 + \cos^2 \frac {\pi x} {3}$$

Take $\log_3$ of both sides to get

$$x^2-6x+11 = 2 + \log_3 \cos^2 \frac {\pi x} {3}$$

The maximum of $\cos^2 \frac {\pi x} {3} = 1$, thus $\log_3 1 = 0$ , so we have

$$x^2 - 6x + 11 = 2$$

Bringing the 2 to the right hand side we have

$$x^2 - 6x + 9 = 0$$

Factoring gives us

$$(x-3)^2 = 0$$

And thus we have $$x = 3$$

Check: $3^{3^2-18+11} = 3^{9-18+11} = 3^2 = 9; \cos^2 \pi = 0; \Rightarrow 9 + 0 = 9 \ \ \checkmark$

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    $\begingroup$ How do you know $\cos^2(\pi x/3)=1$? $\endgroup$ – Tyberius Sep 14 '17 at 1:41
  • $\begingroup$ That's great! So it is also possible to approach with taking logarithm Thank you very much $\endgroup$ – Nariman Zendehrooh Sep 15 '17 at 5:27
  • $\begingroup$ Yes, indeed...it's just the base of the logarithm that matters. $\endgroup$ – bjcolby15 Sep 15 '17 at 10:50

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