Evaluate as a single logarithim

1) 7log16/15 + 5log25/24 + 3log81/80 .

Prove

1) 1/log_a ^abc+ 1/log_b ^abc + 1/log_c ^abc = 1.

Question

Evaluate as a single logarithim
1) 7log16/15 + 5log25/24 + 3log81/80
Prove
1) 1/loga abc + 1/logb
abc + 1/logc abc = 1

Ayush Jha · Accepted Answer

1>We have to evaluate-

7log16/15 + 5log25/24 + 3log81/80

So,

$= \log (\frac{2^{28}}{3^{7} 5^{7}} \frac{5^{10}}{2^{15} 3^{5}} \frac{3^{12}}{2^{12} 5^{3}}) = \log 2$

2>We have,

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

=log_abc ^a+ log_abc ^b+log_abc

=log_abc ^abc

=1

Evaluate as a single logarithim 1) 7log16/15 + 5log25/24 + 3log81/80 . Prove 1) 1/loga abc + 1/logb abc + 1/logc abc = 1.