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I have to calculate the intersections of the two following functions:

(i) f(x) = $3^x$ and $g(x) = 2^{-x}$ (ii) f(x) = $e^{-x}$ and $g(x) = 2e^x$

and I must do a mistake somewhere but I don't know where.

For (i) I simply get $x \log3 = -x \log2$ and hence $x(\log3-\log2) = 0$, so the intersection is at x=0? And for (ii) I get $\log(e^{-x}) = \log(2e^x)$ and hence $-x = \log2+x*\log e$ which then results in $0 = log2$. Where are my mistakes?

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    $\begingroup$ For (i), you're right. For (ii), you made an algebra mistake. $-x = \log 2 + x \log e = \log 2 + x$, so $-2x=\log 2$, or $x=-\frac{1}{2}\log 2$. $\endgroup$
    – mjqxxxx
    Apr 23, 2013 at 23:17

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$(i)$ There is no mistake in your work here: $x = 0$ is correct.

$(ii)$ Your work up to and including this statement is correct: $-x = \log2+x\log e$. I think you made mistakenly canceled the $x$ terms on each side of the equation.

$$ \begin{align} - x & = \log2+x\log e \\ \\ - x& = \log 2 + x \\ \\ -2x & = \log 2 \\ \\ x & = -\frac 12\cdot \log 2 \end{align} $$

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  • $\begingroup$ Very clear description using the OP post! +1 $\endgroup$
    – Amzoti
    Apr 24, 2013 at 0:39

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