Question

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

Solve it Yaar !!!

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

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)}}=

log

15

(3)log

405

(3)

ln(405)

ln(135)

−

log

15

(3)log

405

(3)

ln(15)

ln(5)

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

c

a

±

c

b

=

c

a±b

\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)}}=

log

15

(3)log

405

(3)

ln(405)

ln(135)

−

ln(15)

ln(5)

\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)} Join

ln(405)

ln(135)

−

ln(15)

ln(5)

:

ln(15)ln(405)

ln(15)ln(135)−ln(5)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)}=

log

15

(3)log

405

(3)

ln(15)ln(405)

ln(15)ln(135)−ln(5)ln(405)

=\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)}=

ln(15)ln(405)log

15

(3)log

405

(3)

ln(135)ln(15)−ln(5)ln(405)

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

a

(b)=

ln(a)

ln(b)

\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)}ln(15)log

15

(3)=

ln(e)

ln(15)

⋅

ln(15)

ln(3)

=\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)}=

ln(405)

ln(e)

ln(15)

⋅

ln(15)

ln(3)

log

405

(3)

ln(135)ln(15)−ln(5)ln(405)

=\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)}=

ln(405)

ln(e)

ln(3)

log

405

(3)

ln(135)ln(15)−ln(5)ln(405)

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

a

(b)=

ln(a)

ln(b)

\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)}ln(405)log

405

(3)=

ln(e)

ln(405)

⋅

ln(405)

ln(3)

=\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)}}=

ln(e)

ln(3)

⋅

ln(e)

ln(405)

⋅

ln(405)

ln(3)

ln(135)ln(15)−ln(5)ln(405)

=\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)}}=

ln(e)

ln(3)

⋅

ln(e)

ln(3)

ln(135)ln(15)−ln(5)ln(405)

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

ln(e)

ln(3)

=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)}=

ln(e)

ln(3)

ln(3)

ln(15)ln(135)−ln(5)ln(405)

=\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(15)ln(135)−ln(5)ln(405)

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

ln

2

(3)

ln(15)ln(135)−ln(5)ln(405)

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

=3