Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

Rainbow_Dash

  • 3 years ago

log(7) 1/343

  • This Question is Closed
  1. jrzyby1
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    -3

  2. shandelman
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @jrzyby1 It is against the code of conduct of the site to just give answers ESPECIALLY with zero explanation.

  3. Wired
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 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.

  4. jiteshmeghwal9
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 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\[x=3+(-1)\]\[x=2\]

  5. ParthKohli
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Oopsie. Mistake here. ^

  6. ParthKohli
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \( \color{Black}{\Rightarrow 7^{x} = 7^{-3}}\) \( \color{Black}{\Rightarrow x = -3}\)

  7. ParthKohli
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I think you confused \((7^3)^{-1}\) with \(7^3 \times 7^{-1}\)

  8. jiteshmeghwal9
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i m sorry @ParthKohli . I was really confused

  9. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy