If logπ π = 1/5 solve the equation , π₯ logπ π + (π₯ + 2) logπ π2 = 3
Given: logπ π = 1/5 ...(1)
Now, π₯ logπ π + (π₯ + 2) logπ π2 = 3
β\Rightarrowβ π₯ logπ π + (π₯ + 2).(2 logπ q) = 3 [β΅logam=m.loga\because log a^m = m. log aβ΅logam=m.loga ]
β\Rightarrowβ x5+(x+2)Γ2Γ15=3\frac{x}{5}+(x+2)\times2\times \frac 1 5=35xβ+(x+2)Γ2Γ51β=3 [From eq. (1)]
βx+2x+4=15β3x=11βx=113\Rightarrow x+2x+4=15\\ \Rightarrow 3x=11 \\\Rightarrow x =\frac {11}{3}βx+2x+4=15β3x=11βx=311β
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