Proving Log relation: 2/log(base 9) a - 1/log (base 3) a =

[kimmy_koo51]

2/ log(base 9) a - 1/log (base 3) a = 3/log(base 3) a

 

[skeeter]

hint ...

use the change of base formula

\(\displaystyle \L \log_9{a} = \frac{\log_3{a}}{\log_3{9}} = \frac{\log_3{a}}{2}\)

 

[soroban]

Re: Proving Log relation: 2/log(base 9) a - 1/log (base 3) a

Hello, kimmy_koo51!

Are you familiar with this identity? \(\displaystyle \L\:\log_b(a) \:=\:\frac{1}{\log_a(b)}\)

Prove: \(\displaystyle \L\:\frac{2}{\log_9(a)}\,-\,\frac{1}{\log_3(a)} \:= \:\frac{3}{\log_3(a)}\)

The left side is:

. . . . . . \(\displaystyle \L2\cdot\frac{1}{\log_9(a)} \,- \,\frac{1}{\log_3(a)}\)

. . \(\displaystyle \L=\;2\cdot\log_a(9)\,-\,\log_a(3)\)

. . \(\displaystyle \L=\;\log_a(9^2)\,-\,\log_a(3)\)

. . \(\displaystyle \L=\;\log_a(81)\,-\,\log_a(3)\)

. . \(\displaystyle \L= \;\log_a\left(\frac{81}{3}\right)\)

. . \(\displaystyle \L=\;\log_a(27)\)

. . \(\displaystyle \L=\;\log_a(3^3)\)

. . \(\displaystyle \L=\;3\cdot\log_a(3)\)

. . \(\displaystyle \L=\;\frac{3}{\log_3(a)}\)