Question 823741
{{{7log(6, (x-4))+2log(6, (x+3))-(1/2)log(6, (x))}}}
If these were like terms then we could just add or subtract them to combine them into a single logarithm. (Like logarithmic terms have the same bases and the same arguments.) These, however, are not like terms (same bases, yes, but all three arguments are different: x-4, x+3 and x.)<br>
Fortunately there are three properties of logarithms which provide an laternate way to combine logarithmic terms:<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><li>{{{n*log(a, (p)) = log(a, (p^n))}}}</li></ul>The first two can be used to combine logarithmic terms. (The first one for when the terms have a "+: between them and the second for when there is a "-".) They require that the logs have the same bases and that they have a coefficient of 1.<br>
Our terms all have coefficients that are not 1's. But that is where the third property comes in. It can be used to "move" a coefficient into the argument as its exponent. So that is where we will start: Using the third property to move the coefficients out of the way:
{{{log(6, ((x-4)^7))+log(6, ((x+3)^2))-log(6, (x^(1/2)))}}}
Since 1/2 as an exponent is the same as a square root and since fractional exponents do not always display well on algebra.com, I am going to replace the fractional exponent with a square root before proceeding. (This is not something you need to do.)
{{{log(6, ((x-4)^7))+log(6, ((x+3)^2))-log(6, (sqrt(x)))}}}<br>
Now we will use the first property to combine the first two terms:
{{{log(6, ((x-4)^7 * (x+3)^2))-log(6, (sqrt(x)))}}}
And now we use the second property to combine the remaining terms:
{{{log(6, (((x-4)^7 * (x+3)^2)/sqrt(x)))}}}
This may be an acceptable answer. But we may want to rationalize the denominator:
{{{log(6, ((((x-4)^7 * (x+3)^2)/sqrt(x))(sqrt(x)/sqrt(x))))}}}
{{{log(6, (((x-4)^7 * (x+3)^2 * sqrt(x))/x))}}}