Question

Prove that $\frac{1}{\log_{ab}abc} +\frac{1}{\log_{bc}abc}+\frac{1}{\log_{ca}abc}=2$

Answer 1

Answer:

Step-by-step explanation:

Hi,

Consider L.H.S

= 1/logₐb(abc) + 1/logbc(abc) + 1/logₐc(abc)

Using Change of base property, we can write

logₐb(abc) = log abc/log ab

logbc(abc) = log abc/log bc

logₐc(abc) = log abc/log ac

So, L.H.S can be written as

log ab/log abc + log bc/log abc + log ac/log abc

= ( log ab + log bc + log ac)/log abc

Using Additive Property

log x + log y = log(xy), we get

L.H.S = log(ab*bc*ac)/log abc

= log (abc)²/log (abc)

Using Exponent property log xⁿ = n log x

log (abc)² = 2 log abc

So, L.H.S = 2 log abc/log abc

= 2

= R.H.S

Hope, it helps !

Answer 2

Answer:

Step-by-step explanation:

Concept:

1.Base charging formula:

$log_{a}b=\frac{1}{log_{b}a}$

2.$log_{a}a=1$

3.Product rule of logarithm

$log_{a}M+ log_{a}N= log_{a}(MN)$

Now,

$\frac{1}{log_{ab}abc}+\frac{1}{log_{bc}abc}+\frac{1}{log_{ca}abc}\\\\=log_{abc}(ab)+log_{abc}(bc)+log_{abc}(ca)\\\\=log_{abc}(ab)(bc)(ca)\\\\=log_{abc}(abc)^2\\\\=2 log_{abc}(abc)$

=2 (1)

=2