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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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I wonder how you arrived at a degree 4 polynomial in the first place. There are three integer solutions. – Hagen von Eitzen Sep 13 '12 at 23:11
How do you get a fourth degree equation? If I rewrite it as $\ln\frac{9t+45}{5-t} = \ln((t+3)^2)$ and then remove the logarithms and multiply through by the denominator, what is left is the third degree $(t+3)^2(5-t)-(2t+45)=0$. – Henning Makholm Sep 13 '12 at 23:12
Check your calculations. The equation you get is not the equation I get. – Gerry Myerson Sep 13 '12 at 23:15
@HenningMakholm: that should be $-(9t+45)$ at the end. – Ross Millikan Sep 13 '12 at 23:16
@Ross: Indeed it should. Don't know how I managed to mistype it. – Henning Makholm Sep 13 '12 at 23:18
up vote 0 down vote accepted

$f(t)=t^3-4t^2+25t-35$ 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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True, but irrelevant - the polynomial's wrong. – Gerry Myerson Sep 13 '12 at 23:16
@GerryMyerson: you are right. Maybe the failure of the rational root theorem should be a clue to this. – Ross Millikan Sep 13 '12 at 23:17
@RossMillikan I have been told what the rational root theorem is but not truly what the results when it fails mean. Could you explain more why the failure gives a clue? – yiyi Sep 14 '12 at 0:10
@MaoYiyi: it means there are no rational roots. In this case, the potential roots are $\pm (1,5,7,35),$ so only eight possibilities. If any work, you have one factor of the cubic and can divide by $x-\text{the root}$ to get a quadratic. In this case, none do, so the root is irrational. My comment about it failing is because usually in school problems, the roots are rational, so the failure indicates there may have been an error made earlier. – Ross Millikan Sep 14 '12 at 0:20
@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. – yiyi Sep 14 '12 at 0:31

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