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$f∶\mathbb{R} \to \mathbb{R}, f(x)=6^x-2\cdot3^x-2^x$

How do you solve: $6^{x_1}-2\cdot 3^{x_1}-2^{x_1} = 6^{x_2}-2\cdot 3^{x_2}-2^{x_2}$

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    I've no idea why it doesn't show the right format. – 2017-01-24
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    Hope i don't get thrown with eggs. – 2017-01-24

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We have that $f(x) = 6^x - 2\cdot 3^x - 2^x = (3^x - 1)(2^x - 2) - 2$

Now assume that $f(x_1) = f(x_2)$. Then we have that:

$$(3^{x_1} - 1)(2^{x_1} - 2) = (3^{x_2} - 1)(2^{x_2} - 2)$$

This easily gives us that $f(1) = f(0)$, so the function isn't injective.

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    Isn't it really the same idea, just engineered without proving anything further? – 2017-01-24
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    @Topliner People aren't here to give you full solutions or solve your homework problems. You're always more likely to get small hints, which is why you are asked to share the progress you've made. Anyway try to continue from here and if you encounter a problem we might discuss it. – 2017-01-24
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    Yes,the exercise states show that it isn't injective. On what points exactly? – 2017-01-24
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    @Topliner: Then why did you ask how to show that it ***is*** injective? – 2017-01-24
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    Does it really make a problem? You guys figured it out it isn't injective, whats the fuss? – 2017-01-24
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    @Topliner Nobody made a problem, but you behaved like you expected somebody to give you a full answer. For example it's not hard to see that $f(0) = f(1)$, which can be easily deduced from the equation. – 2017-01-24
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    Thank you. I'm sorry if i made that impression. – 2017-01-24