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-5-4-

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

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