Question

7^2x+3=1 pls help us out​

Answer 1

$7^{2x+3}=1$

$\mathrm{If\:}f\left(x\right)=g\left(x\right)\mathrm{,\:then\:}\ln \left(f\left(x\right)\right)=\ln \left(g\left(x\right)\right)$

$\ln \left(7^{2x+3}\right)=\ln \left(1\right)$

$\begin{aligned}&\text { Apply log rule: } \log _{a}\left(x^{b}\right)=b \cdot \log _{a}(x)\\\\&\ln \left(7^{2 x+3}\right)=(2 x+3) \ln (7)\end{aligned}$

$\left(2x+3\right)\ln \left(7\right)=\ln \left(1\right)$

$\dfrac{\left(2x+3\right)\ln \left(7\right)}{\ln \left(7\right)}=\dfrac{\ln \left(1\right)}{\ln \left(7\right)}$

$2x+3=0$

$2x=-3$

$\Large\boxed{\sf{x=-\dfrac{3}{2}}}$