Question

If $x=\log_{12}6, y=\log_{18}12, z=\log_{24}18$, prove that 1 + xyz = 2yz.

Answer 1

Answer:

Step-by-step explanation:

Hi,

Here we will be using the following properties of  

logarithm:  

logₐb = log b/log a  

nlog a = log aⁿ

log a - log b = log (a/b)

Given x = log₁₂6

which we can rewrite  as x = log 6/log 12

Given y = log₁₈12

which we can rewrite as y = log 12/log 18

Given z = log₂₄18

which we can rewrite as z = log 18/log 24

Consider 2yz - xyz

= 2*(log 12/log 18) *(log 18/log 24) -

  (log 6/log 12)*(log 12/log 18)*(log 18/log 24)

= 2*(log 12/log 24) - log 6/log 24

= 2log 12/log 24) - log 6/log 24

We can write 2log 12 as log 12² = log 144

So, 2yz  - xyz

= log 144/log 24 - log 6/log 24

= (log 144 - log 6)/log 24

= log (144/6)/log 24

= log 24/log 24

= 1

Hence,  2yz - xyz = 1

So xyz + 1 = 2yz

Hence, Proved

Hope, it helps !

Answer 2

Solution :

Given,

$x=\log_{12}6, y=\log_{18}12, z=\log_{24}18$

**********************************
We know the, logarithmic laws:

$\log_{b}a=\log_{c}a\times \log_{b}c$

$\log_{a}a=1$

$logm^{n}=nlogm$

**********************************
LHS = 1 + xyz

= 1 + $log_{12}6×log_{18}12×log_{24}18$

= 1 + $log_{24}6$

= $log_{24}24 + log_{24}6$

= $log_{24}(24×6)$

= $log_{24}(12)^{2}$

= 2$log_{24}12$

= 2$log_{18}12 × log_{24}18$

= $2yz$

= RHS