Solve-the-equation-log-2-x-log-3-x-log-5-x-log-2-x-log-3-x-log-3-x-log-5-x-log-5-x-log-2-x-

Question Number 17386 by ajfour last updated on 05/Jul/17

Solve the equation :

\log_{2} x \log_{3} x \log_{5} x = \log_{2} x \log_{3} x

+ \log_{3} x \log_{5} x

+ \log_{5} x \log_{2} x .

Answered by mrW1 last updated on 05/Jul/17

let u = \ln x

a = \ln 2

b = \ln 3

c = \ln 5

\Rightarrow \frac{u^{3}}{abc} = \frac{u^{2}}{ab} + \frac{u^{2}}{bc} + \frac{u^{2}}{ca}

\Rightarrow u^{3} = (a + b + c) u^{2}

\Rightarrow u^{2} [u - (a + b + c)] = 0

\Rightarrow u = 0

or

\Rightarrow u = a + b + c

\Rightarrow \ln x = 0

\Rightarrow x = 1

\Rightarrow \ln x = \ln 2 + \ln 3 + \ln 5 = \ln (2 \times 3 \times 5)

\Rightarrow x = 2 \times 3 \times 5 = 30

Commented by ajfour last updated on 05/Jul/17

great, you play confident sir!

who else can ?

Commented by 47358857 last updated on 07/Dec/17

Answered by b.e.h.i.8.3.4.17@gmail.com last updated on 06/Jul/17

a b c = a b + b c + c a

\Rightarrow \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 1 \Rightarrow {l o g}_{x}^{2} + {l o g}_{x}^{3} + {l o g}_{x}^{5} = 1

\Rightarrow {l o g}_{x}^{2 \times 3 \times 5} = 1 \Rightarrow {l o g}_{x}^{30} = 1 \Rightarrow x = 30 .

a l s o : x = 1, c a n b e a n a n s w e r .

Commented by ajfour last updated on 08/Jul/17

very nice way, sir .

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