Need Help with This Question If Logba=c and logxb=c, show that logax=1/c^2

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Originally Posted by requal Need Help with This Question If Logba=c and logxb=c, show that logax=1/c^2 Bobak

Originally Posted by requal Need Help with This Question If Logba=c and logxb=c, show that logax=1/c^2 Here is one way. log(b)[a] = c So, a = b^c Or, b = a^(1/c) ----(i) log(x)[b] = c So, b = x^c Or, x = b^(1/c) -----(ii) Plug the b from (i) into (ii), x = [a^(1/c)]^(1/c) x = a^(1 / c^2) -------(iii) Hence, log(a)[x] = log(a)[a^(1 / c^2)] = 1 / c^2

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