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the question : $$a^{\log_{\frac1a}{\frac12}}$$ relevant equation : $$a^ {\log_a(x)} = x$$ $$\log_{c^m} (y) =\frac1m \log_c{(y)}$$ my try at it :
I first changed the base into a by multiplying the log part by $(-1)$. the answer was $a^{ - \log_a{\frac12}}.$ this is equal to $\dfrac{a^1}{\log_a\left(\frac12\right)}$. please help me after that.

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  • $\begingroup$ Hi Esha;I think you need to develop some writing skills before posting a question $\endgroup$ – Learnmore Apr 9 '17 at 2:43
  • $\begingroup$ I am also new but that's what the tour page says $\endgroup$ – Learnmore Apr 9 '17 at 2:43
  • $\begingroup$ Check out the following link for math formatting tips: math.meta.stackexchange.com/questions/5020/… $\endgroup$ – Dave Apr 9 '17 at 2:44
  • $\begingroup$ thanx for showing me this. its really a huge help. do i need to repost this question again? $\endgroup$ – Esha Mukhopadhyay Apr 9 '17 at 2:46
  • $\begingroup$ please pardon me and help me in this question . ill try my best in the future ones $\endgroup$ – Esha Mukhopadhyay Apr 9 '17 at 2:47
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Answer: $2.$

Proof: Let $z = \log_{1/a}(1/2) = \frac{\log_a (1/2)}{\log_a (1/a)} = - \log_a (1/2).$

Then $a^z = a^{- \log_a (1/2)} =$ $\frac {1} {a^{\log_a (1/2)}} =$ $\frac {1} {1/2} = 2.$

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  • $\begingroup$ My edit: You forgot the dollar signs and had 1 missing brace bracket. And I separated the lines. $\endgroup$ – DanielWainfleet Apr 9 '17 at 5:11

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