If 1-loga x-1-logc x-2-logb x- then a-b-c are in

Log_x a+log_x x=2log_xb⇒ ac=b^2 ...

If 1/loga x+1...

Question

If $\frac{1}{l o g_{a} x} + \frac{1}{l o g_{c} x} = \frac{2}{l o g_{b} x}$ , then a,b,c are in

A.P.

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None of these

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Similar questions

[l o g_{b} a l o g_{c} a - l o g_{a} a] + [l o g_{a} b l o g_{c} b - l o g_{b} b] + [l o g_{a} c l o g_{b} c - l o g_{c} c] = 0

a, b, c

(a > 1, b > 1, c > 1)

x

x > 0, x \neq 1

{log}_{a} x, {log}_{b} x, {log}_{c} x

[l o g_{b} a l o g_{c} a - l o g_{a} a] + [l o g_{a} b l o g_{c} b - l o g_{b} b] + [l o g_{a} c l o g_{b} c - l o g_{c} c] = 0, t h e n a b c =

log (\frac{a}{b - c}) = log (\frac{b}{c - a}) = log (\frac{c}{a - b})

a^{a} \times b^{b} \times c^{c} = 1

a, b, c

log a, log b, log c

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