 In this video, we present the solution to question number one for practice exam number one for math 1210 in this one We're asked to expand the logarithm the log base five of the cube root of x squared plus one over x squared minus one So what we're gonna need to use here are lots of logarithms that are useful for expanding logarithmic expressions the first one that comes into mind since The function inside of the logarithm is a fraction of two functions We're gonna first use what we called the second law of logarithms, which tells us That if you take a log any base whatsoever was called base a if you take the log of a divided by b This is equal to the log base a of the numerator Minus the log base a of the denominator. So that's the expression. We're gonna that's the law We're gonna use right here. So that then turns our logarithmic expression We're gonna get the log base five the base doesn't change We're gonna get the cube root of x squared plus one as the first part and then we subtract from that the log base five of x squared minus one and Then so proceeding for what can we do now because if you check the Possible solutions, none of those match up so far. We can expand it a little bit more I want to bring to your attention law number three of logarithms That told us that if you have the log base a of say a to the n power Then exponents inside of the logarithms come out as coefficients And so this becomes n times the log base a of capital a right there And so if we apply the third law to this first logarithmic expression since we're taking the cube root of the whole thing That's the one third power. We can pull out a one third coefficient. And so we get one third times the log base five of x squared plus one and So you'll notice that Most of these logarithms have an x square plus one have some type of coefficient in front of one third That's where that thing's coming from so while some of them seem to be good to go some of them Still a little bit more confusing. What more can we do turns out the first law comes into play as well Because when you look at the denominator x squared minus one that actually factors as a difference of squares x minus one and x plus one Like so in which case then the first law of logarithms comes into play That tells us if we take log base a of a times b This can factor as logged of a plus the log of b And so applying that to the second part right here log base five of x squared minus one that'll become And it should be a minus sign right here Because originally we were subtracting So we're gonna end up with a log base five of x minus one plus a log base five Of x plus one Like so now you'll notice this negative sign actually distributes on the both pieces. So I'm just gonna incorporate it right now Drop this parenthesis whatsoever. So we have a minus sign right here and then this Plus right here actually will be a minus when it's distributed So this actually is our expanded form. We're gonna look for a solution that looks just like this We're gonna get one third log of x square plus one Minus the log of x minus one minus the log of x plus one and so that then leads us to choose option D