Question

Show that:
$\log_{b} a^{5} \cdot \log_{c} b^{3} \cdot \log_{a} c^{7}=105$

Answer 1

BRO I THINK THAT THE QUESTION IS WRONG

THIS IS WHAT I GOT

Answer 2

Step-by-step explanation:

As per the question,

We have been provided the logarithmic equation,

$log_{b} \ a^{5} \cdot log_{c} \ b^{3} \cdot log_{a} \ c^{7} = 105$

Using the property of logarithm:

$log_{y} \ x^{n} = n\cdot log_{y}\ x$

$log_y}\ x=\frac{log\ x}{log\ y}$

Therefore,

$log_{b} \ a^{5} \cdot log_{c} \ b^{3} \cdot log_{a} \ c^{7}$

$5log_{b} \ a \cdot 3log_{c} \ b\cdot 7log_{a} \ c\cdot$

$105 \times \frac{log \ a}{log\ b} \cdot \frac{log \ b}{log\ c}\cdot \frac{log \ c}{log\ a}$

= 105

That is,

LHS = RHS

Hence, proved