Question

if \: a = log_{x}(xyz) \: b = log_{y}(xyz) \: c = log_{c}(xyz) \: then \: prove \: 1 \div a + 1 \div b + 1 \div c = 1ifa=logx​(xyz)b=logy​(xyz)c=logc​(xyz)thenprove1÷a+1÷b+1÷c=1​

Answer 1

$\huge{\tt{\underline{\red{Given}}}}$

a=log$_{x}^{xyz}$

b=log$_{y}^{xyz}$

c= log$_{z}^{xyz}$

To prove:

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

Solution:

We know that

log$_{xyz}^{x}$ = $\frac{1}{a}$

Similarly

$\frac{1}{b}$= log$_{xyz}^{y}$

$\frac{1}{c}$= log$_{xyz}^{z}$

So from this

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

$\implies$ log$_{xyz}^{x}$ + log$_{xyz}^{y}$ + log$_{xyz}^{z}$

Now we have xyz as common base

So as per the property of log

Log$_{xyz}^{x}$ + Log$_{xyz}^{z}$ + Log$_{xyz}^{z}$

$\implies$ log$_{xyz}^{xyz}$

Therefore This equals to 1