Question

If
$4 {}^{a} = 36 {}^{b} = 9 ^{c } \: \: then \: prove \: that \\ \frac{b}{a} + \frac{b}{c } = 1$
Please help

Answer 1

Heya user,

[tex]4^a = 36^b = 9^c \\\\ log 4^a = log 36^b = log9^c \\ \\ Seperating\: two\: of \:them \:, \\ log4^a = log36^b \:\:\: and \:\:\:log 36^b = log9^c \\ a\:log4=b\:log36\:\:\:and\:\:\: b\:log 36 = c\:log9 \\=\ \textgreater \ b/a = \frac{log4}{log36} \:\:\:\:\:\:and\:\:\:\:\:\:b/c = \frac{log9}{log36} \\ =\ \textgreater \ b/a + b/c = \frac{log4 + log9}{log36} = \frac{log4+log9}{log(9*4)} = \frac{log4 + log9}{log4 + log9} = 1[/tex]

And hence, you get your :--> b/a + b/c = 1 --> Proved