Multiplying Logarithms

Using the base-change formula, the expression becomes $\frac{(\log 7)^2}{(\log 35)^2}+\frac{\log 5}{\log 35}\cdot\frac{\log 245}{\log 35}$ .

Combining denominators and simplifying $\log 245$ , we get $\frac{(\log 7)^2+(\log 5)(\log5+2\log7)}{(\log 35)^2}$ .

$=\frac{(\log7)^2+2\log7\log5+(\log5)^2}{(\log35)^2}$ .

$=\frac{(\log7+\log5)^2}{(\log35)^2}=\frac{(\log35)^2}{(\log35)^2}=1$ .

245=5×7^2

$\log_{35}7=a, \log_{35}5=b$

$a^2+b(2a+b)=a^2+2ab+b^2=(a+b)^2$

$(\log_{35}7+ \log_{35}5)^2=1^2=1$