Question 595609
{{{2log((16/15))+log((25/24))-log((32/27))=0}}}
To find out if the left side of the equation is really equal to zero we need to combine/condense them down to a single logarithm. And since the arguments of the logarithms are different, we will need to use the following properties of logs:<ul><li>{{{log(a, (p)) + log(a, (q)) = log(a, (p*q))}}}</li><li>{{{log(a, (p)) - log(a, (q)) = log(a, (p/q))}}}</li></ul>
These properties require that the coefficients of the logs be 1's. Fortunately there is another property of logarithms, {{{q*log(a, (p)) = log(a, (p^q))}}}, which allows us to "move" a coefficient into the argument as its exponent.<br>
So we will start by using the third property to move the 2 in front of the first log:
{{{log(((16/15)^2))+log((25/24))-log((32/27))=0}}}<br>
Now we can start combining terms. The first two logs have a "+" between them so we will use the first property (which also has a "+" between the logs):
{{{log(((16/15)^2*(25/24)))-log((32/27))=0}}}<br>
The remaining logs have a "-" between them so we will use the second property (which also has a "-" between the logs):
{{{log(((16/15)^2*(25/24)/(32/27)))=0}}}<br>
Now we will set about to simplify the big fraction inside the log. Since dividing is the same as multiplying by a reciprocal, I'm going to rewrite the "divide by 32/27" into "multiply by 27/32":
{{{log(((16/15)^2*(25/24)*(27/32)))=0}}}
Next I'm going to start canceling factors that are common to the numerators and denominators. In order to see all the factors I am going to first rewrite the squared fraction without an exponent:
{{{log(((16/15)*(16/15)*(25/24)*(27/32)))=0}}}
and then I'll factor the numerators and denominators:
{{{log(((16/(3*5))*((8*2)/(3*5))*((5*5)/(8*3))*((3*3*3)/(16*2))))=0}}}
Now we can start canceling:
{{{log(((cross(16)/(cross(3)*cross(5)))*((cross(8)*cross(2))/(cross(3)*cross(5)))*((cross(5)*cross(5))/(cross(8)*cross(3)))*((cross(3)*cross(3)*cross(3))/(cross(16)*cross(2)))))=0}}}
As you can see, <i>everything</i> cancels out! So all we are left with is:
{{{log((1)) = 0}}}
And since the zero power of 10 is 1, log(1) is 0:
0 = 0 Check!<br>
Note: The bases of the logarithms could have been <i>any</i> number (as long as all three bases were the same) since the zero power of any base is equal to 1!