given that a,b and c are positive numbers other than 1 , show that log_b a × log_c b × log_a c = 1 hence, evaluate log_(10) 25 × log_2 10 × log


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Question Number 68616 by Rio Michael last updated on 14/Sep/19

$g i v e n t h a t a, b a n d c a r e p o s i t i v e n u m b e r s o t h e r t h a n 1$ $, s h o w t h a t {l o g}_{b} a \times {l o g}_{c} b \times {l o g}_{a} c = 1$ $h e n c e, e v a l u a t e {l o g}_{10} 25 \times {l o g}_{2} 10 \times {l o g}_{5} 4$
	Answered by $@ty@m123 last updated on 14/Sep/19

	$L e t \log_{b} a = x, {l o g}_{c} b = y, {l o g}_{a} c = z$ $\Rightarrow b^{x} = a, c^{y} = b, a^{z} = c$ $N o w,$ $a^{z} = c [\begin{matrix} N o t e : y o u c a n p r o c e e d \\ w i t h b^{x} = a o r c^{y} = b a l s o . \end{matrix}]$ $\Rightarrow {(b^{x})}^{z} = c$ $\Rightarrow {(c^{y})}^{x z} = c$ $\Rightarrow c^{x y z} = c$ $\Rightarrow x y z = 1$ $\Rightarrow {l o g}_{b} a \times {l o g}_{c} b \times {l o g}_{a} c = 1$
	Answered by $@ty@m123 last updated on 14/Sep/19

	$H e n c e$ ${l o g}_{10} 25 \times {l o g}_{2} 10 \times {l o g}_{5} 4$ $= {2 \log}_{10} 5 \times {l o g}_{2} 10 \times 2 {l o g}_{5} 2$ $= 4 (\log_{10} 5 \times {l o g}_{2} 10 \times {l o g}_{5} 2)$ $= 4 \times 1 = 4$
	Commented by Rio Michael last updated on 14/Sep/19

	$t h a n k s s o m u c h$
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