Why apply log in both sides won't work to solve for t in a exponential polynomial function?

Question

I get how to solve this equation in another thread. $$46080e^{−600t}−576000e^{−1500t}+737280e^{−2400t}=0$$ But I didn't understand why it won't work if I apply log in both sides, like: (log being the natural logarithmic)$$log(46080e^{−600t})+log(737280e^{−2400t})=log(576000e^{−1500t})$$ Then:$$log(46080)+log(e^{−600t})+log(737280)+log(e^{−2400t})=log(576000)+log(e^{−1500t})$$ $$log(46080)−600t+log(737280)−2400t=log(576000)−1500t$$ and solve it for t.

Am I missing something?

Answer 1

Your mistake is in this equation:

$ log(46080e^{−600t})+log(737280e^{−2400t})=log(576000e^{−1500t}) $

You should do this instead:

$log(46080e^{−600t}+737280e^{−2400t})=log(576000e^{−1500t})$

Answer 2

Put $u=\exp(-300t)$

Then $46080u^2-576000u^5+737280u^8=0$

$46080=(2^{10})(3^2)5\\576000=(2^9)(3^2)(5^3)\\737280=(2^{14}(3^2)(5)$

So let's divide by $23040u^2=(2^9)(3^2)(5)u^2$

$$2-25u^3+32u^6=0$$

$\Delta=25^2-8\times 32=369=(3^2)(41)$

Thus $u^3=\dfrac{25\pm3\sqrt{41}}{64}\qquad$ [ both roots are positive ]

And then we get $t=-\dfrac{\ln(u^3)}{900}$

t1=0.000411051405
t2=0.002669602730

---

Regarding your question, the logarithm transforms products into sums: $\ln(ab)=\ln(a)+\ln(b)$

Since here you have a sum on exponentials, then it is no use to take the logarithm of it, this won't help simplifying the expression.

Answer 3

$46080 e^{-600 t}-576000 e^{-1500 t}+737280 e^{-2400 t}=0$

Divide both sides by $e^{-600 t}$

$46080 -576000 e^{-900 t}+737280 e^{-1800 t}=0$

set $e^{-900 t}=w\to e^{-1800 t}=w^2$

$737280w^2-576000w+46080=0$

$$w_1= \frac{1}{64} \left(25-3 \sqrt{41}\right),\;w_2= \frac{1}{64} \left(25+3 \sqrt{41}\right)$$

reset $w=e^{-900 t}$

we have

$$e^{-900 t}=\frac{1}{64} \left(25-3 \sqrt{41}\right)\to -900t=\log\left(\frac{1}{64} \left(25-3 \sqrt{41}\right)\right)\\ t_1=-\frac{1}{900}\log\left(\frac{1}{64} \left(25-3 \sqrt{41}\right)\right)\approx 0.0026696$$

$$t_2\approx 0.000411051$$

Hope this is useful