#### nicholaskong100

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- Thread starter nicholaskong100
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Hi Nicholas. There's no formula. We use what we know about logs and exponents (i.e., definitions and properties -- for writing/manipulating expressions).Is there a name of a formula

Do you know how to switch back and forth between exponential form and logarithmic form? That could be one approach.

Let the right-hand side be y.

[math]\log_{2^3}(x) = y[/math]

Switch to exponential form, and then solve for the variable in the exponent.

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(That's the "exponential form" Otis mentioned. Basically, [imath]a^x[/imath] and [imath]log_a(x)[/imath] are "inverse functions".)

Saying that [imath]y= log_{2^3}(x)[/imath] is the same as saying that [imath]x= (2^3)^y= 2^{3y}[/imath].

Going back, then, [imath]3y= log_2(x)[/imath] or [imath]y=\frac{1}{3}log_2(x)[/imath].

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Thanks for following up with that. I like this approach because, even if a student doesn't realize that the definition can be used with exponential form to go back directly, they ought to possess prior experience of solving for y by taking logs and using the property to get y out of the exponent position.Saying that [imath]y= log_{2^3}(x)[/imath] is the same as saying that [imath]x= (2^3)^y= 2^{3y}[/imath].

Going back, then, [imath]3y= log_2(x)[/imath] ...

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Thanks, oh. I didn't know. Now, I understand it now.

(That's the "exponential form" Otis mentioned. Basically, [imath]a^x[/imath] and [imath]log_a(x)[/imath] are "inverse functions".)

Saying that [imath]y= log_{2^3}(x)[/imath] is the same as saying that [imath]x= (2^3)^y= 2^{3y}[/imath].

Going back, then, [imath]3y= log_2(x)[/imath] or [imath]y=\frac{1}{3}log_2(x)[/imath].

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Thanks I find out how to turn log into exponent and it's much easier to understand now. Thank you for the keywords, Otis.Hi Nicholas. There's no formula. We use what we know about logs and exponents (i.e., definitions and properties -- for writing/manipulating expressions).

Do you know how to switch back and forth between exponential form and logarithmic form? That could be one approach.

Let the right-hand side be y.

[math]\log_{2^3}(x) = y[/math]

Switch to exponential form, and then solve for the variable in the exponent.