The laws of logarithms are inverse operations of exponential functions that allow simplification of logarithmic expressions through three key rules: the product rule (log_b(xy) = log_b(x) + log_b(y)), the quotient rule (log_b(x/y) = log_b(x) - log_b(y)), and the power rule (log_b(x^n) = n·log_b(x)). These laws enable the expansion, condensation, and evaluation of logarithmic expressions by converting multiplication to addition, division to subtraction, and exponents to coefficients.
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Section 8 3Added:
now we're going to look at the laws of logarithms they're going to help us undo exponentials so you got to remember these are are the inverse of everything that we've done with these exponential functions so when we make our laws they come out of the exponent laws that we've used up to this point okay so we add exponents if we have two exponents with two bases so when we do this as an inverse basically if we see two logs added together that means we can multiply bring it down from from two logs to one by using multiplication we can also expand them if we have log base two of two numbers multiplied together we can actually make this into two log base twos with those numbers separated and it's separated by addition quotient rule we use subtraction so we would take our exponents we'd subtract them works the same way with logs except in this case if you see log base 2 with division you can break it into two separate log base twos using subtraction and the last one is power rule the power rule has if you had one base two exponents you multiply the exponents in this case multiplication is going to happen if you have one base and you have a log uh the log number inside with the exponent then you're going to multiply the 4 by the log so the exponent comes up front exponent times log this one here should say m there's a mistake on that so it should say p times log base c of m because the m stays the p becomes what you multiply by okay so let's try a few of these they're pretty simple and straightforward as long as you just kind of look at them first so the very first one a if i look at that i can see that i'm using division so according this if i have a log single log and i'm dividing what's inside of it because this xy is inside of the logarithm i can expand this out by making it two logs so i can go log base 6 of x and because it's division i subtract minus log base 6 of y that's how you do a b b you can see that there's a power inside of here it's x times y to the n and they're both raised to the power of one half so the first thing i would do is get that one half out front so this would become log base 5 of x y to the power of one-half so that got rid of that radical sign i know this is a power of a power so i can bring the one half out front and then go log base 5 of x y and at this moment i've got a single log but i've got multiplication happening inside of it if i want to break this up i make it into addition so i would go one half and then i'd need a big bracket because the half is in front of all of this xy i'm going to split it into two pieces i have to make sure the one half goes to both log base 5 of x plus because of multiplication log base 5 of y and if i want to continue expanding that i would put the one half through the brackets and go one half log base five of x plus one half log base 5 of the y now it's fully expanded part c there's a lot going on in there i think i'm going to do the division first so when i look at that i can see that it's 9 divided by the cube root of x squared so i'm going to make that division sign into a subtraction sign by giving the log to both the top and bottom of the fraction so here's the numerator log base 3 of 9 minus the denominator gets a log base 3 as well log base 3 of the cube root of x squared and this is basically saying 3 raised to what power equals nine that's what that's what that's saying so you can just make this a two two minus and i'm going to turn this uh radical sign into an exponent so it would be log base 3 of x squared to the power of 1 3.
this then becomes two minus log base three of x to the two-thirds because those have to be multiplied together one base two exponents and finally i would take the two thirds out front and i just get log base three of x and that should clean that up as much as it can the last one part d maybe i will erase this to see if i can make myself a little bit of room part d is actually way easier than it looks because part d what you're going to do is just hold on part d you've got two pieces so the first piece is the top of the fraction the numerator so i'm going to take that on the log base 7 of x to the fifth y that's my first piece and the second piece is the denominator subtract log base 7 of z to the power of one half i just made the square root into one half now i have to expand the first piece because the numerator has multiplication in it x to the fifth plus y so i expand that out by using addition log base 7 of x to the power of 5 plus log base 7 of y subtract log base 7 of z to the power of one half now i'm going to bring down all the exponents out front so it would be 5 log base 7 of x plus log base 7 of y minus one half log base 7 of z and you're done problem wasn't too bad it looked worse than it was to move to the next page it says use the laws of logarithms to simplify and evaluate each expression so this one in the first one here part a you can see this says log base 3 of 9 times root 3.
when i see multiplication i know i can change it to addition by just giving the log base 3 to both the 9 and the root 3. so that's what i'm going to do equals log base 3 of 9 plus log base three of root three which is three to the one half this says three raised to the power of what equals nine three squared is nine so two is the answer so log log base three of nine the other one i have to bring out one half i bring it down front that's a power of a power and i get log base 3 of 3. let's say 3 to the power of what equals 3 well that's just 1. so this is just 2 to the 2 plus 1 half and if i get a common denominator i multiply the top and bottom by two i get four plus one over two which is five over two that's the answer to that question part b i recommend doing these step by step so you can see there's subtraction and subtraction again so you're going to divide twice but what happens is when people try to do it all at once a lot of times they'll make a mistake in the way they write it so just recommend going through and going log base five and dividing the first two numbers one thousand divided by four and just carrying this one on for the ride until we can get to him log base 5 of 1000 divided by 4 is 250.
and so now we have 2 again and division so i would go log base 5 of 250 divided by 2 which is 125 this is saying 5 to the power of what equals 125.
so the answer is 3.
that's how i rewrite this to know how to get my answer um c hopefully i can get this done in this much room first thing i would do is i would put all the exponents back on top of the numbers here so like 2 log base 3 of 6 becomes log base 3 of 6 squared and then subtract log base 3 of the square root of 64 plus log base 3 of 2.
so now i'll just multiply them out log base 3 of 36 minus log base 3 of square root of 64 is 8 plus log base 3 of 2.
i'm going to start putting them together now so i got log base 3 of 36 divided by 8 plus log base 3 of 2.
36 divided by 8 does not give me a nice number so i'm going to keep going with this by 36 divided by 8 and then it says plus log base 3 of 2 36 divided by 8 times 2 is what that would become so now i go log base 3 36 divided by 8 times 2 and now i can see that i can reduce 2 goes into 8 4 times that will leave me with log base 3 of 36 divided by four which is nine log base three of nine which the answer is just two there we've got those guys down the last two last two are pretty long but let's take a try so if i start doing this what i would do is probably what's in the brackets first so and i can take this 4 back up as an exponent so this becomes log base 3 of x to the power 4.
subtract one-half times and if i put these back together because it's addition i would i would multiply so i'd get log base 3 of x times x to the fifth because there's a 5 there out in front so he has to also go back up all right that gives me log base three of x to the fourth minus one half and in here x to the one times x to the fifth we add our exponents so this would be log base 3 of x to the sixth now take the first one leave it alone we take this one half back and we end up with log base 3 of x to the sixth all to the power of one half which is actually six divided by two is three log base three of x to power 4 minus log is 3 of x to the power of 3.
this is division so i can put them together i just go x to the 4th divided by x cubed x to the fourth divided by x cubed we subtract exponents 4 minus 3 is 1 and our answer is log base 3 of x the second one you know just a as an aside this answer i have to remember with these that there are restrictions on them x must be greater than zero because if i put log base three of zero in my calculator i'll get an answer of error we have to remember that when we're dealing with a log function it's kind of shaped like this and there's an asymptote at zero that it can't cross so our we have do have restrictions i won't be asking for them often but i do need to do you to remember that you do restrict domains on some of these okay looks like this is log base 2 of x squared minus 9 minus log base 2 of x squared minus x minus 6. so it looks like we can just make it into a single log base 2 by using division this is true because it says subtraction let's be x squared minus nine over x minus three times x plus two so i got a little carried away there i actually factored that two things that multiply to negative six add to negative one one on top can also be factored x squared minus nine is x plus three and x minus three is the difference of squares and that's all over x minus 3 times x plus 2.
and you can see by that that you can reduce that fraction and you will end up with log base 2 of x plus 3 over x plus two so that one's done now you can work on your assignment
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