Here's the question you clicked on:
Rainbow_Dash
log(7) 1/343
@jrzyby1 It is against the code of conduct of the site to just give answers ESPECIALLY with zero explanation.
\[\Large b^{y}=x\] \[\Large \log_{b}{x}=y\] \[\Large \log_{7}{\frac{1}{343}}=y\] \[\Large 7^{y}=\frac{1}{343}=343^{-1}\] \[\Large 7^{y}=343^{-1}\] Solve.
\[\log_{7}{1 \over 343}=x \]can change it into exponential form as\[7^x=\frac{1}{343}\]& u can write it as\[7^x=343^{-1}\]\[7^x=(7^3)^{-1}\]now u know that if the bases are equal then powers are also equal so u can write the equation as\[x=3+(-1)\]\[x=2\]
Oopsie. Mistake here. ^
\( \color{Black}{\Rightarrow 7^{x} = 7^{-3}}\) \( \color{Black}{\Rightarrow x = -3}\)
I think you confused \((7^3)^{-1}\) with \(7^3 \times 7^{-1}\)
i m sorry @ParthKohli . I was really confused