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I have these two equations: $$ ln(9t+45)-ln(5-t)= ln(t+3)^{2}$$

$$and$$

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

When I solve the first equation I get that $$t=0, t=3, t=-4$$

However, when I solve the second equation I get $$t=0, t=3$$ But arent these two equations the same? Shouldn't the second equations also give t=-4 as a solution as well? Because on the right-hand side for the second equation I have only applied the power rule for ln.

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  • $\begingroup$ @ HåkanMjölk I guess the mistake must have been that for the first equation, you squared $\ln(t+3)$ whereas for the second, you doubled it. Please notify if you had done it right. Calculation mistakes can arise anyhow if not careful. $\endgroup$
    – Spectre
    Sep 21, 2021 at 9:04
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    $\begingroup$ I guess you mixed up these: $$ \ln (a^2) = 2 \ln (a) $$ $$ \ln (b) ^2 = (\ln b)^2 \neq 2 \ln b $$ $\endgroup$
    – Matti P.
    Sep 21, 2021 at 9:05
  • $\begingroup$ I wonder what you were taught as "the power rule for logarithm". Normally it says, if $a>0$ and $b$ are real numbers, then $\log(a^b)=b\log a$. Note the bit $a>0$. This means that, for $t+3\le 0$ you cannot apply that rule. More generally, whichever rule you want to apply, there may be conditions attached to that rule, which are just as important as the rule itself. $\endgroup$ Sep 21, 2021 at 9:48
  • $\begingroup$ A formatting tip: write \ln instead of ln. $\endgroup$ Sep 21, 2021 at 9:58

1 Answer 1

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The problem here is the following: you think that $$\ln (A^2) = 2 \ln (A)$$ is always true. But the truth is that $$\mbox{IF } A>0 \qquad \mbox{ THEN } \ln (A^2) = 2 \ln (A)$$

In your particular case the first equation has different conditions from the second one. Indeed $$\ln (9t+45) - \ln (5-t) = 2 \ln (t+3)$$ requires that $\ln(t+3)$ exists, i.e. $t > -3$, while $$\ln (9t+45) - \ln (5-t) = \ln (t+3)^2$$ does not (because $(t+3)^2>0$ is automatically satisfied). Since $-4<-3$ you have that $t=-4$ is a solution of the second equation, while it is not a solution of the first equation.

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