if-log-6-45-x-log-18-24-y-then-log-12-25-in-terms-of-x-y-

Question Number 25593 by behi.8.3.4.17@gmail.com last updated on 11/Dec/17

i f : {l o g}_{6}^{45} = x, {l o g}_{18}^{24} = y

t h e n : l o g_{12}^{25} = ? (in terms of x, y)

Answered by Rasheed.Sindhi last updated on 14/Dec/17

\log_{6} 45 = x \Rightarrow 6^{x} = 45 \Rightarrow 2^{x} . 3^{x} = 3^{2} . 5 \dots \dots \dots . (i)

\log_{18} 24 = y \Rightarrow 18^{y} = 24 \Rightarrow 2^{y} . 3^{2 y} = 2^{3} . 3 \dots \dots . . (ii)

\log_{12} 25 = z (say) \Rightarrow 12^{z} = 25 \Rightarrow 2^{2 z} . 3^{z} = 5^{2} \dots . . (iii)

Determining z in terms of x, y .

(i) \times (ii) \times (iii) :

(2^{x} . 3^{x}) (2^{y} . 3^{2 y}) (2^{2 z} . 3^{z}) = (3^{2} . 5) (2^{3} . 3) (5^{2})

2^{x + y + 2 z} . 3^{x + 2 y + z} = 2^{3} . 3^{3} . 5^{3}

2^{x + y} . 2^{2 z} . 3^{x + 2 y} . 3^{z} = 2^{3} . 3^{3} . 5^{3}

{(2^{2})}^{z} {(3)}^{z} = \frac{2^{3} . 3^{3} . 5^{3}}{2^{x + y} . 3^{x + 2 y}}

12^{z} = 2^{3 - x - y} . 3^{3 - x - 2 y} . 5^{3}

\log (12^{z}) = \log (2^{3 - x - y} . 3^{3 - x - 2 y} . 5^{3})

\log_{12} 25 = \frac{\log (2^{3 - x - y} . 3^{3 - x - 2 y} . 5^{3})}{\log (12)}

Or

\log_{12} 25 = \log_{12} (2^{3 - x - y} . 3^{3 - x - 2 y} . 5^{3})

Commented by Rasheed.Sindhi last updated on 14/Dec/17

The answer has been corrected .