Given, loga\(x\) = \(\frac{1}{α}\), logb\(x\) = \(\frac{1}{β}\), lo

1.	Given, loga\(x\) = \(\frac{1}{α}\), logb\(x\) = \(\frac{1}{β}\), logc\(x\) = \(\frac{1}{γ}\), then logabc\(x\) equals :(a) αβγ (b) \(\frac{1}{αβγ}\)(c) α + β + γ(d) \(\frac{1}{α + β+γ}\)
Answer» (d) \(\frac{1}{α + β+γ}\) α = \(\frac{1}{log_ax}\), β = \(\frac{1}{log_bx}\), γ = \(\frac{1}{log_cx}\) ⇒ α = log_xa, β = log_xb, γ = log_xc ⇒ α + β + γ = log_xa + log_xb + log_xc = log_x(abc) ⇒ \(\frac{1}{α + β+γ}\) = \(\frac{1}{log_x(abc)}\) = log_abcx.

Given, loga\(x\) = \(\frac{1}{α}\), logb\(x\) = \(\frac{1}{β}\), logc\(x\) = \(\frac{1}{γ}\), then logabc\(x\) equals :(a) αβγ (b) \(\frac{1}{αβγ}\)(c) α + β + γ(d) \(\frac{1}{α + β+γ}\)

Answer»

(d) \(\frac{1}{α + β+γ}\)

α = \(\frac{1}{log_ax}\), β = \(\frac{1}{log_bx}\), γ = \(\frac{1}{log_cx}\)

⇒ α = log_xa, β = log_xb, γ = log_xc

⇒ α + β + γ = log_xa + log_xb + log_xc = log_x(abc)

⇒ \(\frac{1}{α + β+γ}\) = \(\frac{1}{log_x(abc)}\) = log_abcx.