pre-calculus-if-log-48-72-log-54-12-k-then-log-12-27-

Question Number 135285 by mnjuly1970 last updated on 11/Mar/21

p r e - c a l c u l u s

i f {l o g}_{48}^{72} + {l o g}_{54}^{12} = k

t h e n {l o g}_{12}^{27} = ? ? ?

Answered by bobhans last updated on 12/Mar/21

\frac{\ln 72}{\ln 48} + \frac{\ln 12}{\ln 54} = k

\frac{\ln 6 + \ln 12}{\ln 6 + \ln 8} + \frac{\ln 12}{\ln 9 + \ln 6} = k

(1) \ln 6 = \ln 2 + \ln 3 = a + b

(2) \frac{3 \ln 2 + 2 \ln 3}{4 \ln 2 + \ln 3} + \frac{2 \ln 2 + \ln 3}{3 \ln 3 + \ln 2} = k

\Rightarrow \frac{3 a + 2 b}{4 a + b} + \frac{2 a + b}{a + 3 b} = k

\Rightarrow (3 a + 2 b) (a + 3 b) + (4 a + b) (2 a + b) = k (4 a + b) (a + 3 b)

\Rightarrow 3 a^{2} + 11 a b + 6 b^{2} + 8 a^{2} + 6 a b + b^{2} = 4 {k a}^{2} + 13 k a b + 3 {k b}^{2}

\Rightarrow (11 - 4 k) a^{2} + (17 b - 13 k b) a + 7 b^{2} - 3 {k b}^{2} = 0

\Rightarrow a = \frac{13 k b - 17 b + \sqrt{{(17 b - 13 k b)}^{2} - 4 (11 - 4 k) ((7 b^{2} - 3 {k b}^{2})}}{22 - 8 k}

\log_{12} (27) = \frac{\ln 27}{\ln 12} = \frac{3 \ln 3}{2 \ln 2 + \ln 3} = \frac{3 b}{2 a + b}

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