171. $7^{2 x}-48 \cdot 7^{x}=49$ 

 

171. $7^{2 x}-48 \cdot 7^{x}=49$
Solution. Let $7^{x}=y$, we get the quadratic equation $y^{2}-$ $-48 y-49=0$. Let's solve it. Here $a=1, b=-48, c=-49 ; \quad D=b^{2}-$ $-4 a c=(-48)^{2}-4 \cdot 1(-49)=2304+196=2500 ; \sqrt{D}=50$. Using the formula $y_{1,2}=\frac{-b \pm \sqrt{D}}{2 a}$, we find

$$
y_{1}=\frac{48-50}{2}=\frac{-2}{2}=-1 ; \quad y_{2}=\frac{48+50}{2}=\frac{98}{2}=49
$$

Since $7^{x}=y$, then $7^{x}=-1$ (this equality is impossible, as the exponential function can only take positive values); $7^{x}=49 ; 7^{x}=7^{2}$, i.e., $x=2$. Thus, we get the answer: $x=2$.