Question

$\frac{log_3135\:}{log_{15}3}-\frac{log_35\:}{log_{405}3}$

Crown for best answer

Answer 1

♣ Qᴜᴇꜱᴛɪᴏɴ :

$\large\boxed{\sf{\dfrac{log_3135\:}{log_{15}3}-\dfrac{log_35\:}{log_{405}3}}}$

♣ ᴀɴꜱᴡᴇʀ :

$\boxed{\sf{\dfrac{\log _3\left(135\right)}{\log _{15}\left(3\right)}-\dfrac{\log _3\left(5\right)}{\log _{405}\left(3\right)}=\dfrac{\ln \left(15\right)\ln \left(135\right)-\ln \left(5\right)\ln \left(405\right)}{\ln ^2\left(3\right)}\quad \left(\mathrm{Decimal:\quad }\:3\right)}}$

♣ ᴄᴀʟᴄᴜʟᴀᴛɪᴏɴꜱ :

$\text { Least Common Multiplier of } \log _{15}(3), \log _{405}(3): \log _{15}(3) \log _{405}(3)$

Adjust Fractions based on L.C.M

$\sf{=\dfrac{\dfrac{\ln \left(135\right)}{\ln \left(405\right)}}{\log _{15}\left(3\right)\log _{405}\left(3\right)}-\dfrac{\dfrac{\ln \left(5\right)}{\ln \left(15\right)}}{\log _{15}\left(3\right)\log _{405}\left(3\right)}}$

$\mathrm{Since\:the\:denominators\:are\:equal,\:combine\:the\:fractions}:\quad \dfrac{a}{c}\pm \dfrac{b}{c}=\dfrac{a\pm \:b}{c}$

$\sf{=\dfrac{\dfrac{\ln \left(135\right)}{\ln \left(405\right)}-\dfrac{\ln \left(5\right)}{\ln \left(15\right)}}{\log _{15}\left(3\right)\log _{405}\left(3\right)}}$

$\text { Join } \dfrac{\ln (135)}{\ln (405)}-\dfrac{\ln (5)}{\ln (15)}: \quad \dfrac{\ln (15) \ln (135)-\ln (5) \ln (405)}{\ln (15) \ln (405)}$

$=\dfrac{\dfrac{\ln \left(15\right)\ln \left(135\right)-\ln \left(5\right)\ln \left(405\right)}{\ln \left(15\right)\ln \left(405\right)}}{\log _{15}\left(3\right)\log _{405}\left(3\right)}$

$=\dfrac{\ln \left(135\right)\ln \left(15\right)-\ln \left(5\right)\ln \left(405\right)}{\ln \left(15\right)\ln \left(405\right)\log _{15}\left(3\right)\log _{405}\left(3\right)}$

$\mathrm{Apply\:log\:rule}:\quad \log _a\left(b\right)=\dfrac{\ln \left(b\right)}{\ln \left(a\right)}$

$\ln \left(15\right)\log _{15}\left(3\right)=\dfrac{\ln \left(15\right)}{\ln \left(e\right)}\cdot \dfrac{\ln \left(3\right)}{\ln \left(15\right)}$

$=\dfrac{\ln \left(135\right)\ln \left(15\right)-\ln \left(5\right)\ln \left(405\right)}{\ln \left(405\right)\dfrac{\ln \left(15\right)}{\ln \left(e\right)}\cdot \dfrac{\ln \left(3\right)}{\ln \left(15\right)}\log _{405}\left(3\right)}$

$=\dfrac{\ln \left(135\right)\ln \left(15\right)-\ln \left(5\right)\ln \left(405\right)}{\ln \left(405\right)\dfrac{\ln \left(3\right)}{\ln \left(e\right)}\log _{405}\left(3\right)}$

$\mathrm{Apply\:log\:rule}:\quad \log _a\left(b\right)=\dfrac{\ln \left(b\right)}{\ln \left(a\right)}$

$\ln \left(405\right)\log _{405}\left(3\right)=\dfrac{\ln \left(405\right)}{\ln \left(e\right)}\cdot \dfrac{\ln \left(3\right)}{\ln \left(405\right)}$

$=\dfrac{\ln \left(135\right)\ln \left(15\right)-\ln \left(5\right)\ln \left(405\right)}{\dfrac{\ln \left(3\right)}{\ln \left(e\right)}\cdot \dfrac{\ln \left(405\right)}{\ln \left(e\right)}\cdot \dfrac{\ln \left(3\right)}{\ln \left(405\right)}}$

$=\dfrac{\ln \left(135\right)\ln \left(15\right)-\ln \left(5\right)\ln \left(405\right)}{\dfrac{\ln \left(3\right)}{\ln \left(e\right)}\cdot \dfrac{\ln \left(3\right)}{\ln \left(e\right)}}$

$\dfrac{\ln (3)}{\ln (e)}=\ln (3)$

$=\dfrac{\ln \left(15\right)\ln \left(135\right)-\ln \left(5\right)\ln \left(405\right)}{\dfrac{\ln \left(3\right)}{\ln \left(e\right)}\ln \left(3\right)}$

$=\dfrac{\ln \left(15\right)\ln \left(135\right)-\ln \left(5\right)\ln \left(405\right)}{\ln \left(3\right)\ln \left(3\right)}$

$\ln (3) \ln (3)=\ln ^{2}(3)$

$=\dfrac{\ln \left(15\right)\ln \left(135\right)-\ln \left(5\right)\ln \left(405\right)}{\ln ^2\left(3\right)}$

$\huge\boxed{\sf{=3}}$

Answer 2

log2 : log with base 2 ,

“ ” log3 : log with base 3 ,

“ ” log4 : log with base 4 ,

“ ” log5 : log with base 5 ,

“ ” log6 : log with base 6 ,

···

The base can be any positive real number

not necessarily an integer

. Actually,

from a logical standpoint, working with just one of those logs suffices, because all

others are easily recovered from it. Though the choice of such ‘representative log’ is

arbitrary, for a variety of reasons, the consensus most natural choice is the one with

“base e”:

“ ” ln = loge : natural log .

I will tell you the reason why. But for the time being I treat all of them equally