log-6-15-a-log-12-18-b-find-log-25-24-in-terms-of-a-and-b-

Question Number 126430 by TITA last updated on 20/Dec/20

\log_{6}^{15} = a \log_{12}^{18} = b find \log_{25}^{24}

in terms of a and b

Commented by TITA last updated on 20/Dec/20

please help

Answered by mr W last updated on 20/Dec/20

l e t \log 2 = x, \log 3 = y

\log_{6} 15 = \frac{\log 15}{\log 6} = \frac{\log 3 + \log 10 - \log 2}{\log 3 + \log 2}

= \frac{\log 3 + 1 - \log 2}{\log 3 + \log 2} = \frac{y + 1 - x}{y + x} = a

\Rightarrow (1 + a) x - (1 - a) y = 1 \dots (i)

\log_{12} 18 = \frac{\log 18}{\log 12} = \frac{\log 2 + 2 \log 3}{\log 3 + 2 \log 2}

= \frac{x + 2 y}{y + 2 x} = b

\Rightarrow (1 - 2 b) x + (2 - b) y = 0 \dots (i i)

f r o m (i) a n d (i i) w e g e t

x = \frac{b - 2}{(1 + a) (b - 2) - (1 - a) (1 - 2 b)}

y = \frac{1 - 2 b}{(1 + a) (b - 2) - (1 - a) (1 - 2 b)}

\log_{25} 24 = \frac{\log 24}{\log 25} = \frac{\log 3 + 3 \log 2}{2 - 2 \log 2} = \frac{3 x + y}{2 (1 - x)}

= \frac{3 (b - 2) + (1 - 2 b)}{2 [(1 + a) (b - 2) - (1 - a) (1 - 2 b) - b + 2]}

\Rightarrow \frac{b - 5}{2 (2 b - a - a b - 1)}

Commented by Dwaipayan Shikari last updated on 20/Dec/20

W e h a v e 9 ° C i n K o l k a t a

Commented by MJS_new last updated on 20/Dec/20

lol I^{'} m baking cookies so I^{'} m not as fast as you \dots

Commented by mr W last updated on 20/Dec/20

i^{'} m j u s t e a t i n g c o o k i e s w i t h G l \ddot{u h w e i n},

s o i m u s t b e f a s t e r \dots

Commented by MJS_new last updated on 20/Dec/20

I made Elisenlebkuchen, Vanillekipferl &

Zimtsterne \dots season greetings from Vienna!

Commented by Dwaipayan Shikari last updated on 20/Dec/20

W h a t i s t h e w e a t h e r i n V i e n n a s i r ? M a y b e c o l d

Commented by Ar Brandon last updated on 20/Dec/20

��Mr W and Mr MJS

Commented by MJS_new last updated on 20/Dec/20

{it}^{'} s now 5 : 40 p . m ., sunset was 4 : 02 p . m .

we have + 3 ° C and mist / light rain

Commented by talminator2856791 last updated on 22/Dec/20

in south africa it is around 20 ° in morning now

Answered by MJS_new last updated on 20/Dec/20

a = \frac{\ln 15}{\ln 6} = \frac{\ln 3 + \ln 5}{\ln 2 + \ln 3}

b = \frac{\ln 18}{\ln 12} = \frac{\ln 2 + 2 \ln 3}{2 \ln 2 + \ln 3}

let \ln 2 = p; \ln 3 = q; \ln 5 = r

a = \frac{q + r}{p + q} \Rightarrow p = \frac{- q a + q + r}{a}

b = \frac{p + 2 q}{2 p + q} = \frac{- q a - q - r}{q a - 2 q - 2 r} \Rightarrow q = \frac{(2 b - 1) r}{a b + a - 2 b + 1}

\Rightarrow p = \frac{(2 - b) r}{a b + a - 2 b + 1}

we need to know \frac{\ln 24}{\ln 25} = \frac{3 \ln 2 + \ln 2}{2 \ln 5} = \frac{3 p + q}{2 r}

and inserting we get

\frac{5 - b}{2 (a b + a - 2 b + 1)}

Answered by Dwaipayan Shikari last updated on 20/Dec/20

\frac{l o g 15}{l o g 6} = a \Rightarrow l o g (5) + l o g (3) = a l o g (3) + a l o g (2) \to (a)

\frac{l o g (18)}{l o g (12)} = b \Rightarrow 2 l o g (3) + l o g (2) = b l o g (3) + 2 b l o g (2) \Rightarrow \frac{l o g 3}{l o g 2} = \frac{2 b - 1}{2 - b}

F r o m (a) l o g (5) = (a - 1) l o g 3 + a l o g (2) = l o g (2) (\frac{(a - 1) (2 b - 1)}{2 - b} + a)

\frac{l o g (8) + l o g (3)}{2 l o g (5)} = \frac{3 l o g (2) + l o g (2) (\frac{2 b - 1}{2 - b})}{2 l o g (2) (\frac{2 a b - 2 b - a + 1 + 2 a - a b}{2 - b})} = \frac{5 - b}{2 (a b + a - 2 b + 1)}

Commented by Dwaipayan Shikari last updated on 20/Dec/20

S l o w e s t p o s s i b l e a n s w e r