```Question 629415
{{{(1/5)(3log((x+1))+2log((x-1))-log((7)))}}}
There are two ways to condense/combine logarithms:<ul><li>Adding or subtracting them if they are like terms. (Like logarithmic terms have the same bases and same arguments.)</li><li>Using either of the following properties of logarithms:<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 logs have the same bases and coefficients of 1.</li></ul>The logs in your expression are all base 10 logs. They have different arguments however so we will not be able to add or subtract them.<br>
Two of your logarithms do not have coefficients of 1 so, at the moment, we cannot use the properties to combine the terms. 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. By using this property we can create the coefficients of 1 we need to use the other two properties to combine the terms.
Using this 3rd property on the first two logs we get:
{{{(1/5)(log(((x+1)^3))+log(((x-1)^2))-log((7)))}}}<br>
We can now start using the first two properties to combine the logs. The first two logs have a "+" between them so we will use the first property (since it also has a "+" between the logs):
{{{(1/5)(log(((x+1)^3(x-1)^2))-log((7)))}}}
Now we will use the second property because of the "-" between the remaining logs:
{{{(1/5)(log((((x+1)^3(x-1)^2)/7)))}}}
This may the the desired "single quantity". Or perhaps we should use the 3rd property to move the 1/5 into the argument, too:
{{{log(((((x+1)^3(x-1)^2)/7)^(1/5)))}}}
This may be the desired answer. Or since 1/5 as an exponent means 5th root:
{{{log((root(5, ((x+1)^3(x-1)^2)/7)))}}}
```