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$$log_a (x) = y$$

$$\Rightarrow x = a^{y}$$

Similarly, if $$antilog_a (y) = x$$ How will it be written in exponential form?

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    If $x=a^y$, and $x=\operatorname{antilog}_a(y)$, then by transitivity... – 2017-01-07
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    See here: http://www.mathcaptain.com/algebra/inverse-logarithm.html – 2017-01-07
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    anti$\log_a y = a^y$ – 2017-01-07

3 Answers 3

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By definition, the anti-log is the inverse of the log:

$$\operatorname{antilog}_a(y)=x$$

Take the log of both sides:

$$\log_a(\operatorname{antilog}_a(y))=\log_a(x)$$

They cancel, and we are left with

$$y=\log_a(x)$$

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    Same thinking I was just typing the answer. – 2017-01-07
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Since the anti-log is the inverse of the log: $${antilog}_a(y)=x$$ Taking log both the sides: $$\log_a(x)=\log_a({antilog}_a(y))$$ The $\log$ and $anti\log$ will cross each other $$y=\log_a(x)$$ Hence, $$a^y=x$$

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If $$antilog_a(y) = x $$ $$\Rightarrow a^{y} = x $$

credit: John & J. M from the comments

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    But you used the answer by the "Simple Art" and "Harsh Kumar" – 2017-01-07