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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Harsh Kumar
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Hi Esha;I think you need to develop some writing skills before posting a question – Learnmore Apr 09 '17 at 02:43
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I am also new but that's what the tour page says – Learnmore Apr 09 '17 at 02:43
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Check out the following link for math formatting tips: https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – Dave Apr 09 '17 at 02:44
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thanx for showing me this. its really a huge help. do i need to repost this question again? – Esha Mukhopadhyay Apr 09 '17 at 02:46
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please pardon me and help me in this question . ill try my best in the future ones – Esha Mukhopadhyay Apr 09 '17 at 02:47
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Do you mean $a^{\log_{\frac{1}{a}}\frac{1}{2}}$? – Juniven Acapulco Apr 09 '17 at 02:55
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thanx for formatting it. i m finally able to figure out the answer. i was making a silly mistake all this while. – Esha Mukhopadhyay Apr 09 '17 at 03:28
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what should i start the formatting with? i mean to say do i hav to write a head, body and end just like in programming? – Esha Mukhopadhyay Apr 09 '17 at 13:08
1 Answers
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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 $a^z = a^{- \log_a (1/2)} =$ $\frac {1} {a^{\log_a (1/2)}} =$ $\frac {1} {1/2} = 2.$
DanielWainfleet
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Raoul Ohio
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My edit: You forgot the dollar signs and had 1 missing brace bracket. And I separated the lines. – DanielWainfleet Apr 09 '17 at 05:11