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# Using the logarithmic power rule

CCSS.Math:

## Video transcript

we're asked to simplify log base 5 of X to the 3rd and once again we're just going to rewrite this in a different way you could argue whether it's going to be more simple or not and the logarithm property that I'm guessing that we should use for this example right here it's the property if I take log log base X of let me pick some more letters here of log base X of Y to the Z power that this is the same thing as Z times log base X of Y so this is a logarithm property if I'm taking the logarithm of a given base of something to a power I could take that power out front I could take that power out front and multiply that times the log of the base of just the Y in this case so if we apply this property over here and in a second once I do this problem we'll talk about why this actually makes a lot of sense and comes straight out of exponent properties but if we just apply that over here we get log base 5 of X to the third well this is the exponent right over here that's the same thing as Z so that's going to be the same thing as let me just in a different color that 3 is the same thing we can put it out front that's the same thing as 3 times the logarithm base 5 of X and we're done this is just another way of writing it using this property and so you could argue that this is a what maybe this is a simplification because you took the exponent outside of the logarithm and you're multiplying the logarithm by that number now now with that out of the way let's think about why that actually makes sense so let's say that we let's say that we know let's say that we know that I'll just pick some arbitrary letters here let's say that we know that a to the B power is equal to C and so if we know that that's written as an exponential equation if we wanted to write the same truth as a logarithmic logarithmic equation we would say logarithm base a of C is equal to B to what power do I have to raise a to get C I raise it to the B power a to the B power is equal to C fair enough now let's take both sides of this equation right over here and raise it to the dthe power so let me make it so let's raise take both sides of this equation and raise it to the D power instead of doing it in place I'm just going to rewrite it over here so I wrote the original equation a to the B is equal to C which is just rewriting this statement but let me take both sides of this to the D power and I should be consistent I'll use all capital letters so this should be a B actually let's say it's what I'm losing all lowercase letters so this is a lowercase e so let me write it this way a to the so I I'm going to raise this to the D power and I'm going to raise this to the D power obviously if these two things are equal to each other if I raise both sides of the same power the Equality is still going to hold now what's interesting over here what's interesting over here is we can now say what we can do is we can use our what we know about exponent property to say look if I have a to the B power and then I raise that to the D power our exponent properties say that this is the same thing this is equal to this is equal to a to the BD power this is equal to a to the BD let me write it here this is let me do that in a different color of our use that green this right over here using what we know about exponent properties this is the same thing as a to the BD power so we have a to the BD power is equal to C to the D power and now this this exponential equation if we write it as a logarithmic equation we would say we would say log base a of C to the D power C to the D power is equal to BD is equal to BD what power do I have to raise a to to get to C to the dthe power to get to this I have to raise it to the BD power but what do we know that that B is we already know that B is this thing right over here so if we substitute this in for B and we can rewrite this as DB we get we get logarithm base a of C to the D power is equal to BD or you could also call that D I'm just if you switch the order and so that's equal to D times B B is just log log base a of C so there you have it we just derived this property log base a of C to the dthe that's the same thing as D times log base a of C which we applied right over here