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lo g π π ( x − π ) = lo g π ( e π ) , x = ?
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2 solutions
Ok, thanks
Just for the sake of obtaining knowledge, why were you able go from the original equation to the second step?
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By the properties of logarithms we have lo g a n b = n 1 lo g a b .
A beautiful solution! Liked it and learnt a lot. Thank you!
We can apply exponentiation with base π π anthe ecuation gets x − π = π π l o g π ( e π ) = > x = e π π + π = > x = e + π I didn't create this website, but I'm sure the one or ones that did it were not thinking about contravening each other like babies, please, try to control yourselves. Maths are not for this
We have: \log_\sqrt [π]{π} (x - π) = \log_{π} (e^π) \\ \implies \dfrac {1}{\frac {1}{\pi}} \log_{\pi}{(x-\pi)}= \log_{π} (e^π) \\ \implies \log_{\pi}{(x-\pi)^{\pi}}=\log_{\pi}{e^{\pi}} \\ \implies (x-\pi)^{\pi}=e^{\pi} \\ \implies x-\pi=e \\ \therefore x=e+\pi