Question 365314
{{{log(b, (8)) + 5(log(b, (7 + x)) - (1/2)log(b, (6 - x)))}}}
There are three properties of logarithms we will need for this problem:<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>{{{q*log(a, (p)) = log(a, (p^q))}}}</li></ul>
All of these properties can be used in both directions. But for this problem we will using them from left to right. The third property allows us to move a coefficient into the argument as an exponent. We will use it to move the (1/2) in front of the last log into its argument:
{{{log(b, (8)) + 5(log(b, (7 + x)) - log(b, ((6 - x)^(1/2))))}}}
Since 1/2 as an exponent means square root, I'll replace the last argument with a square root:
{{{log(b, (8)) + 5(log(b, (7 + x)) - log(b, (sqrt(6 - x))))}}}
Next, inside the parentheses we have two logarithms of the same base with coefficients of 1 and with a "-" between them. This is a job for the second property:
{{{log(b, (8)) + 5(log(b, ((7 + x)/sqrt(6 - x))))}}}
or
{{{log(b, (8)) + 5*log(b, ((7 + x)/sqrt(6 - x)))}}}
Next we can use the third property again to move the 5 from in front of the second log:
{{{log(b, (8)) + log(b, (((7 + x)/sqrt(6 - x))^5))}}}
And last, with two logs of the same base with coefficients of 1 and a "+" between them, we can use the first property to combine them:
{{{log(b, (8*((7 + x)/sqrt(6 - x))^5))}}}
which is a single logarithm with a coefficient of 1. (NOTE: You do have a square root in a denominator. So you may need to rationalize the denominator. I'll leave that up to you.)