log x - log(x+14) = -1 can anyone help me solve please
Rewrite the left side as a logarithm: log (x/(x+14)
Becuase log (A/B) = log A - log B
So your equation is log(x/(x+14)) = -1
To solve that, write that equation in exponential form..and you'll almost be done.
\(\bf log x - log(x+14) = -1 \implies log_{10} x - log_{10}() = -1 \\ \quad \\ log_{10}\left(\cfrac{x}{x+14}\right)=-1\\ \quad \\ \textit{log cancellation rule of}\quad a^{log_ax}=x\\ \quad \\ \large 10^{log_{10}\left(\frac{x}{x+14}\right)}=10^{-1}\implies \cfrac{x}{x+14}=10^{-1}\)
hmm I have a typo.... .. well. \(\bf log( x) - log(x+14) = -1 \implies log_{10} x - log_{10}(x+14) = -1 \\ \quad \\ log_{10}\left(\cfrac{x}{x+14}\right)=-1\\ \quad \\ \textit{log cancellation rule of}\quad a^{log_ax}=x\\ \quad \\ \large 10^{log_{10}\left(\frac{x}{x+14}\right)}=10^{-1}\implies \cfrac{x}{x+14}=10^{-1}\)
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