(1/25)quad=5^x+5 eching

$$( \frac { 1 } { 25 } ) \quad = 5 ^ { x + 5 }$$

x uchun yechish

$$x=-7$$

Logaritm ifodasidan foydalanish qadamlari

Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.

$$5^{x+5}=\frac{1}{25}$$

Tenglamaning ikkala tarafiga tegishli logaritmni chiqarish.

$$\log(5^{x+5})=\log(\frac{1}{25})$$

Darajaga ko'tarigan logaritm raqami raqam logaritmining darajasidir.

$$\left(x+5\right)\log(5)=\log(\frac{1}{25})$$

Ikki tarafini \log(5) ga bo‘ling.

$$x+5=\frac{\log(\frac{1}{25})}{\log(5)}$$

Asosiy tenglamani almashtirish orqali \frac{\log(a)}{\log(b)}=\log_{b}\left(a\right).

$$x+5=\log_{5}\left(\frac{1}{25}\right)$$

Tenglamaning ikkala tarafidan 5 ni ayirish.

$$x=-2-5$$

x uchun yechish (complex solution)

$$x=\frac{i\times 2\pi n_{1}}{\ln(5)}-7$$

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