5^x2-5x+6=1 eching

$${ 5 }^{ x2-5x+6 } =1$$

x uchun yechish

$$x=\frac{x_{2}+6}{5}$$

Logaritm ifodasidan foydalanish qadamlari

Tenglamani yechish uchun eksponent va logaritmlarning qoidalaridan foydalanish.

$$5^{-5x+x_{2}+6}=1$$

Tenglamaning ikkala tarafiga tegishli logaritmni chiqarish.

$$\log(5^{-5x+x_{2}+6})=\log(1)$$

Darajaga ko'tarigan logaritm raqami raqam logaritmining darajasidir.

$$\left(-5x+x_{2}+6\right)\log(5)=\log(1)$$

Ikki tarafini \log(5) ga bo‘ling.

$$-5x+x_{2}+6=\frac{\log(1)}{\log(5)}$$

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

$$-5x+x_{2}+6=\log_{5}\left(1\right)$$

Tenglamaning ikkala tarafidan x_{2}+6 ni ayirish.

$$-5x=-\left(x_{2}+6\right)$$

Ikki tarafini -5 ga bo‘ling.

$$x=-\frac{x_{2}+6}{-5}$$

x_2 uchun yechish

$$x_{2}=5x-6$$

Logaritm ifodasidan foydalanish qadamlari

Tenglamani yechish uchun eksponent va logaritmlarning qoidalaridan foydalanish.

$$5^{x_{2}+6-5x}=1$$

Tenglamaning ikkala tarafiga tegishli logaritmni chiqarish.

$$\log(5^{x_{2}+6-5x})=\log(1)$$

Darajaga ko'tarigan logaritm raqami raqam logaritmining darajasidir.

$$\left(x_{2}+6-5x\right)\log(5)=\log(1)$$

Ikki tarafini \log(5) ga bo‘ling.

$$x_{2}+6-5x=\frac{\log(1)}{\log(5)}$$

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

$$x_{2}+6-5x=\log_{5}\left(1\right)$$

Tenglamaning ikkala tarafidan -5x+6 ni ayirish.

$$x_{2}=-\left(6-5x\right)$$

x uchun yechish (complex solution)

$$x=-\frac{2\pi n_{1}i}{5\ln(5)}+\frac{x_{2}}{5}+\frac{6}{5}$$

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

x_2 uchun yechish (complex solution)

$$x_{2}=\frac{2\pi n_{1}i}{\ln(5)}+5x-6$$

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