# Logarithm question

#### Poliagapitos

##### New member
$\log_{a^n}b=\frac{1}{\log_ba^n}=\frac{1}{n \cdot \log_ba}=\frac{1}{n} \cdot \frac{1}{\log_ba}=\frac{1}{n}\cdot \log_ab$

#### Otis

##### Elite Member
Is there a name of a formula
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.

$\log_{2^3}(x) = y$
Switch to exponential form, and then solve for the variable in the exponent. • #### pka

##### Elite Member
The standard change of base is [imath]\log_b(a)=\dfrac{\log(a)}{\log(b)}[/imath]

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• Subhotosh Khan

#### HallsofIvy

##### Elite Member
Do you understand that [imath]y= log_a(x)[/imath] is the same as [imath]x= a^y[/imath]?
(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].

#### Otis

##### Elite Member
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] ...
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. • Jomo

##### New member
Do you understand that [imath]y= log_a(x)[/imath] is the same as [imath]x= a^y[/imath]?
(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].
Thanks, oh. I didn't know. Now, I understand it now.

##### New member
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.

$\log_{2^3}(x) = y$
Switch to exponential form, and then solve for the variable in the exponent. Thanks I find out how to turn log into exponent and it's much easier to understand now. Thank you for the keywords, Otis.