So here we have an equation that is very, very centered around having mastered the basic rules of logarithms. So let's start by recognizing what rules of calculation we're going to use. So we see here that we have the logarithm. And by the way, here we see that all logarithms are base 10. So here I'm just going to write down all the rules of calculation and so on, even with base 10, even though it applies to pretty much all bases. So here we have the logarithm of a fraction and the logarithm of a fraction a over B. It can be written as the logarithm of a minus the logarithm of b. And here we have the logarithm of the square root of x. And what we can do is we can write the square root of x as x raised to the power of half. And we also have a rule of calculation to deal with that here because this is a power. So if you have the logarithm of a power, so a raised to the power of b, we can write it as B times the logarithm of a. So then we have a way to deal with it. And here we have the logarithm of a product, so we have the logarithm. to a times B which we can write as the logarithm of a plus the logarithm of b so there we have the arithmetic rules we are going to use and Let's just start writing out eh here so that we get Eh let's just say expanded all those parenthetical expressions so Here we have the logarithm of 10 Sorry 100 over X We can then write that as the logarithm of 100 minus the logarithm of x so that was the one we take color code and Here we have the logarithm of the root of X or in other words the logarithm of x raised to the power of half so the logarithm of x raised to the power of half We can then write that as half times the logarithm of x so this we can write as half times the logarithm of x Then we can take the last term here which is minus and when we are going to expand something that has a minus in front of it it is very important that we introduce a parenthesis here just so that we don't get the sign wrong so the logarithm of 10 the logarithm of 10 plus the logarithm of x in and and again Here we have a power which is the argument of the logarithm And then can we expand it using that rule here So let's just do it right away it becomes 2 logarithm of x and this should then be equal to 7 Hal I apologize for my voice By the way it's a bit of a cold So let's just write a new line where we expand that parenthesis in orange here so we have the logarithm of 100 minus the logarithm of x We have plus one and a half times the logarithm of x and we then have minus the logarithm of 10 and we have minus 2 logarithm of x and again this should be equal to 7 half It's not coming out very nicely yet eh What I like to do when we have eh an expression that is repeated in several places like lgx does here is to introduce a substitution a very big fan of substitution so that I want to introduce Then U is equal to LG X because it will make the expression a little nicer and a little easier to keep track of So let's do that then we have the logarithm of 100 and now I'll drop the color code By the way we just have LG x which we now write as U then we have plus one half g u minus the logarithm of 10 minus 2 UU is equal to 7 half and the only logarithm We are left with here It is the logarithm of 100 and the logarithm of 10 and note that this is the logarithm with base 10 So when it says here what is the logarithm of 100 it just means What do you have to raise 10 to the power of 10 to get 100 and that gives that x must be equal to 2 for Bl 10 to the second is clearly 100 so this here it is only 2 similarly we can also do for that logarithm here What number do you have to raise 10 to the power of 10 to get 10 It is clearly 1 You have to raise 10 to the power of 1 to get 10 so this here is only 1 so then we can write it even a little easier Then we have 2 minus U plus one half U minus 1 minus 2 U is equal to 7 half And then we can just start isolating eh Or we can start combining The terms that have U in itself so we have minus U minus 2 U Then we are down to minus 3 U and then we get plus half a U So then we are down to minus 2 and a half U so we can write it as minus 2.5 U and then we have Then we have included the one we have included and we have included it then it says 2 minus 1 which is 1 so we have plus 1 and it should equal 7 half so then it starts to look a little nicer Eh what we are after now is to isolate u on one side because it is U which has something to do with X And it is X that we are going to solve for so we solve it here as a regular equation for u first so we subtract 1 on both sides maybe start introducing another color too and then we have minus 2.5 U equals 7 half minus 1 and 1 we can write as 2 twos Then we have 7 minus 2 which is 5 then it becomes 5 twos 5 twos and 2.5 that is also exactly that same as 5 twos what is written here is minus 5 twos times U equals 5 twos And then we can divide by 5 twos on both sides and then we actually get the effect just cross it off against it piece by piece it's a word eh anyway so we're just left with 1 here so then we have minus U equals 1 and if we times by minus 1 on both sides then we have U equals minus 1 So then we're almost there we're just going to reverse because it's actually X we're going to solve for so then we reverse the substitution we did which really just made the notation very easy so then we have that the logarithm of x is equal to minus 1 and that's perfectly fine what we can do now is use further logarithm rules we can take and put both sides of the equal sign as an exponent of the number 10 and we use the number 10 because this is the logarithm with base 10 so we have 10 raised to the logarithm of x which Then will be equal to 10 raised to the minus f and 10 raised to the logarithm of base 10 for X it just becomes X so we have X = equal to 10 i minus f which can also be written as 1 tiel so there we have the answer to the problem X is equal to 1 ti