Question

Evaluate the following:
i) $\frac{\log_{4} 7}{\log_{4} 5} \times \frac{\log_{9} 5}{\log_{9} 7}$
ii)$\log_{25} \sqrt{5} \times \log_{3}\frac{1}{9}$

Answer 1

$< marque > hope < marque >$

Answer 2

Answer:

(i) 1

(ii) $\frac{-1}{2}$

Step-by-step explanation:

In the question,

We have been provided the equations,

To evaluate the equation we get,

(i) $\frac{log_{4}7}{log_{4}5}\times \frac{log_{9}5}{log_{9}7}$

Now, from the properties of the logarithms we know that,

$log_{a}b=\frac{logb}{loga}$

So, using the same we get,

$\frac{log_{4}7}{log_{4}5}\times \frac{log_{9}5}{log_{9}7}=\frac{(\frac{log7}{log4})}{(\frac{log5}{log4})}\times \frac{(\frac{log5}{log9})}{(\frac{log7}{log9})}\\=\frac{log7}{log5}\times \frac{log5}{log7}\\=\frac{1}{1}\\=1$

Therefore, on evaluating this equation we get,

$\frac{log_{4}7}{log_{4}5}\times \frac{log_{9}5}{log_{9}7}=1$

(ii) $log_{25}\sqrt{5}\times log_{3}\frac{1}{9}$

Now, from the properties of the logarithms we know that,

$loga^{b}=b.loga\\and,\\log\frac{a}{b}=loga-logb$

So, using the same we get,

$log_{25}\sqrt{5}\times log_{3}\frac{1}{9}=log_{5^{2}}(5)^{\frac{1}{2}}\times (log_{3}1-log_{3}(3)^{2})\\=(\frac{1}{2}\times \frac{1}{2})(log_{5}5)\times (-2log_{3}3)\\=\frac{1}{4}\times (-2)\\=\frac{-1}{2}$

Therefore, on evaluating this equation we get,

$log_{25}\sqrt{5}\times log_{3}\frac{1}{9}=\frac{-1}{2}$