Réitigh 28=36e^-k10

$$28 = 36 e ^ { - k 10 }$$

Réitigh do k_10.

$$k_{10}=\ln(\frac{9}{7})\approx 0.251314428$$

Céimeanna ag Baint Úsáid as Sainiú an Logartaim

Roinn an dá thaobh faoi 36.

$$\frac{28}{36}=e^{-k_{10}}$$

Laghdaigh an codán \frac{28}{36} chuig na téarmaí is ísle trí 4 a bhaint agus a chealú.

$$\frac{7}{9}=e^{-k_{10}}$$

Athraigh na taobhanna ionas go mbeidh na téarmaí inathraitheacha ar fad ar an taobh clé.

$$e^{-k_{10}}=\frac{7}{9}$$

Ghlac logartam an dá thaobh den chothromóid.

$$\log(e^{-k_{10}})=\log(\frac{7}{9})$$

Is ionann logartam uimhreacha a ardaítear go cumhacht agus an chumhacht méadaithe faoi logartam na huimhreach.

$$-k_{10}\log(e)=\log(\frac{7}{9})$$

Roinn an dá thaobh faoi \log(e).

$$-k_{10}=\frac{\log(\frac{7}{9})}{\log(e)}$$

Leis an bhfoirmle athrú boinn \frac{\log(a)}{\log(b)}=\log_{b}\left(a\right).

$$-k_{10}=\log_{e}\left(\frac{7}{9}\right)$$

Roinn an dá thaobh faoi -1.

$$k_{10}=\frac{\ln(\frac{7}{9})}{-1}$$

Réitigh do k_10. (complex solution)

$$k_{10}=-2\pi n_{1}i+\ln(\frac{9}{7})$$

$$n_{1}\in \mathrm{Z}$$