Name: Logarithms in Five Minutes
Uploaded: 2019-07-19
Duration: 5 min 43 s
Description: Logarithms are a very interesting and useful concept used in maths, physics and many more areas. Let's gather all the important information about them in a f...

Question

If

x = \log_{c} b + \log_{b} c, y = \log_{a} c + \log_{c} a, z = \log_{b} a + \log_{a} b

, then show that

x^{2} + y^{2} + z^{2} - 4 = x y z

.

Accepted Answer

Given that $x = \log_{c} b + \log_{b} c, y = \log_{a} c + \log_{c} a, z = \log_{b} a + \log_{a} b$

Now, $x^{2} + y^{2} + z^{2} - x y z$

$= {(\log_{c} b + \log_{b} c)}^{2} + {(\log_{a} c + \log_{c} a)}^{2} + {(\log_{b} a + \log_{a} b)}^{2} - [(\log_{c} b + \log_{b} c) (\log_{a} c + \log_{c} a) (\log_{b} a + \log_{a} b)]$

$= {(\log_{c} b)}^{2} + {(\log_{b} c)}^{2} + 2 \log_{c} b \log_{b} c + {(\log_{a} c)}^{2} + {(\log_{c} a)}^{2} + 2 \log_{a} c \log_{c} a + {(\log_{b} a)}^{2} + {(\log_{a} b)}^{2} + 2 \log_{b} a \log_{a} b - [(\log_{c} b + \log_{b} c) (\log_{a} c \log_{b} a + \log_{a} c \log_{a} b + \log_{c} a \log_{b} a + \log_{a} b \log_{c} a)]$

$= {(\log_{c} b)}^{2} + {(\log_{b} c)}^{2} + (2 \times 1) + {(\log_{a} c)}^{2} + {(\log_{c} a)}^{2} + (2 \times 1) + {(\log_{b} a)}^{2} + {(\log_{a} b)}^{2} + (2 \times 1) - [(\log_{c} b + \log_{b} c) (\log_{b} c + \log_{a} c \log_{a} b + \log_{c} a \log_{b} a + \log_{c} b)]$

$= {(\log_{c} b)}^{2} + {(\log_{b} c)}^{2} + {(\log_{a} c)}^{2} + {(\log_{c} a)}^{2} + {(\log_{b} a)}^{2} + {(\log_{a} b)}^{2} + 6 - [(\log_{c} b \log_{b} c + \log_{c} b \log_{a} c \log_{a} b + \log_{c} a \log_{b} a \log_{c} b + \log_{c} b \log_{c} b) + (\log_{b} c \log_{b} c + \log_{b} c \log_{a} b \log_{a} c + \log_{c} a \log_{b} c \log_{b} a + \log_{b} c \log_{c} b)]$

$= {(\log_{c} b)}^{2} + {(\log_{b} c)}^{2} + {(\log_{a} c)}^{2} + {(\log_{c} a)}^{2} + {(\log_{b} a)}^{2} + {(\log_{a} b)}^{2} + 6 - [(1 + \log_{a} b \log_{a} b + \log_{c} a \log_{c} a + \log_{c} b \log_{c} b) + (\log_{b} c \log_{b} c + \log_{a} c \log_{a} c + \log_{b} a \log_{b} a + 1)]$

$= {(\log_{c} b)}^{2} + {(\log_{b} c)}^{2} + {(\log_{a} c)}^{2} + {(\log_{c} a)}^{2} + {(\log_{b} a)}^{2} + {(\log_{a} b)}^{2} + 6 - [{(\log_{a} b)}^{2} + {(\log_{c} a)}^{2} + {(\log_{c} b)}^{2} + {(\log_{b} c)}^{2} + {(\log_{a} c)}^{2} + {(\log_{b} a)}^{2} + 2]$

$= {(\log_{c} b)}^{2} + {(\log_{b} c)}^{2} + {(\log_{a} c)}^{2} + {(\log_{c} a)}^{2} + {(\log_{b} a)}^{2} + {(\log_{a} b)}^{2} + 6 - {(\log_{a} b)}^{2} - {(\log_{c} a)}^{2} - {(\log_{c} b)}^{2} - {(\log_{b} c)}^{2} - {(\log_{a} c)}^{2} - {(\log_{b} a)}^{2} - 2$

$= 4$

Thus, $x^{2} + y^{2} + z^{2} - x y z = 4$

$\Rightarrow x^{2} + y^{2} + z^{2} - 4 = x y z$

Hence, proved.

If $x = \log_{c} b + \log_{b} c, y = \log_{a} c + \log_{c} a, z = \log_{b} a + \log_{a} b$ , then show that $x^{2} + y^{2} + z^{2} - 4 = x y z$ .

Important Questions on Logarithm

Learn and Prepare for any exam you want !

If x=logcb+logbc, y=logac+logca, z=logba+logab, then show that x2+y2+z2-4=xyz.

Important Questions on Logarithm

Learn and Prepare for any exam you want !

If $x = \log_{c} b + \log_{b} c, y = \log_{a} c + \log_{c} a, z = \log_{b} a + \log_{a} b$ , then show that $x^{2} + y^{2} + z^{2} - 4 = x y z$ .