# Why dosent my equations give the same result when though they are the same

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.

• @ 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. Sep 21, 2021 at 9:04
• I guess you mixed up these: $$\ln (a^2) = 2 \ln (a)$$ $$\ln (b) ^2 = (\ln b)^2 \neq 2 \ln b$$ Sep 21, 2021 at 9:05
• 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. Sep 21, 2021 at 9:48
• A formatting tip: write \ln instead of ln. Sep 21, 2021 at 9:58

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.