A question in logarithm v2

Question

If $x=1+log_a \,bc$, $y=1+log_b \, ca $ and $z=1+log_c \, ab $ , then prove that $xyz= xy+xz+yz $.

My attempt:

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I wish u can view the mathjax properly, bcoz I can't view it well. — Wrichik Basu, May 13 '17 at 16:06
We need $$\dfrac1x+\dfrac1y+\dfrac1z=1$$ Now $$x=1+\log_abc=\log_aa+\log_bc=\log_a(abc)\implies\dfrac1x=?$$ — lab bhattacharjee, May 13 '17 at 16:07

score 1 · Answer 1 · answered May 13 '17 at 16:10

1

$x=\log_a(abc)$, $y=\log_b(abc)$ and $z=\log_c (abc)$

$$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\log_{abc}a+\log_{abc}b+\log_{abc}c=\log_{abc}(abc)=1$$

The result follows.

answered May 13 '17 at 16:10

CY Aries

23,393

score 1 · Accepted Answer · answered May 13 '17 at 16:10

$x=1+\log_abc=\log_aa+\log_abc=\log_aabc\Rightarrow\dfrac{1}{x}=\log_{abc}a,\\y=1+\log_bca=\log_bb+\log_bca=\log_babc\Rightarrow\dfrac{1}{y}=\log_{abc}b,\\z=1+\log_cab=\log_cc+\log_cab=\log_cabc\Rightarrow\dfrac{1}{z}=\log_{abc}c.$

$\text{Now,}$

$\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\log_{abc}a+\log_{abc}b+\log_{abc}c\\\Rightarrow\dfrac{xy+yz+zx}{xyz}=\log_{abc}{abc}=1\\\Rightarrow xy+yz+zx=xyz\hspace{10pt}\cdots\text{Proved}$

A question in logarithm v2

2 Answers2