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shandelman Group TitleBest ResponseYou've already chosen the best response.0
@jrzyby1 It is against the code of conduct of the site to just give answers ESPECIALLY with zero explanation.
 2 years ago

Wired Group TitleBest ResponseYou've already chosen the best response.1
\[\Large b^{y}=x\] \[\Large \log_{b}{x}=y\] \[\Large \log_{7}{\frac{1}{343}}=y\] \[\Large 7^{y}=\frac{1}{343}=343^{1}\] \[\Large 7^{y}=343^{1}\] Solve.
 2 years ago

jiteshmeghwal9 Group TitleBest ResponseYou've already chosen the best response.0
\[\log_{7}{1 \over 343}=x \]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\[x=3+(1)\]\[x=2\]
 2 years ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.0
Oopsie. Mistake here. ^
 2 years ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.0
\( \color{Black}{\Rightarrow 7^{x} = 7^{3}}\) \( \color{Black}{\Rightarrow x = 3}\)
 2 years ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.0
I think you confused \((7^3)^{1}\) with \(7^3 \times 7^{1}\)
 2 years ago

jiteshmeghwal9 Group TitleBest ResponseYou've already chosen the best response.0
i m sorry @ParthKohli . I was really confused
 2 years ago
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