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I'm to solve the equation

$\ln(9t+45) - \ln(5-t) = \ln(t+3)^2$

After some work I arrive at this:

$t(t^3-4t^2+25t-35) = 0$

which clearly shows that $0$ is a root for $t$. This is also clear when testing:

$\ln(0+45) - \ln(5-0) = \ln(0+3)^2 \iff \ln\frac{45}{5} = \ln9$

But how do I know this is the only solution? I can't see any way of factorising $t(t^3-4t^2+25t-35)$ any further, and I can't see any obvious roots. I can't just assume $t=0$ is the only solution, can I?

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    Hm, yes, thanks. Still not sure how I managed to get it so wrong. Thanks!2012-09-14

1 Answers 1

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$f(t)=t^3-4t^2+25t-35$ is an odd order polynomial, so will have at least one real root and may have three. You can try the rational root theorem, but that is no help here. From the fact that $f(0)=-35 \lt 0$ and $f(t) \to +\infty$ as $t \to +\infty$ you know there is a positive root. In fact, $f(2)=7$, so the root is between $0$ and $2$. Then you can take the derivative, see that it is always positive, and know there is only one real root. I basically got this by plotting the graph.

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    @RossMillikan oh, i thought you were talking about the problem on a more complex level, about how the failure would have some extra special meaning not typically taught in basic calc class.2012-09-14