How to show that \(\displaystyle-{{\log}_{{b}}{x}}={{\log}_{{\frac{{{1}}}{{{b}}}}}{x}}\)

Cash Duncan

Cash Duncan

Answered question

2022-04-08

How to show that logbx=log1bx

Answer & Explanation

regulerenes4w

regulerenes4w

Beginner2022-04-09Added 10 answers

we get
log1bxln(x)ln(1b)=ln(x)ln(1)ln(b)=ln(x)ln(b)
izvozna39g0

izvozna39g0

Beginner2022-04-10Added 9 answers

log1bx=logbxlogb1b=logbx-1=-logbx

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