Question 404103
4log(5x+3)-2log(x+15)-1/2log(5x+2)
These are not like logarithmic terms. Like logarithmic terms have logarithms of the same base <i>and same argument.</i> Your logartihms are all base 10 logarithms but the arguments are all different. Since only like logarithmic terms can be subtracted, we cannot subtract these terms.<br>
There is another way. There are two 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 only require that the logarithms have the same base <i>and coefficients of 1!</i> Your are of the same base but the coefficients are not 1's. Fortunately there is a third property of logarithms, {{{q*log(a, (p)) = log(a, (p^q))}}}, that allows us to move a coefficient of a logarithm into the argument as its exponent. So we'll use this third property to move the coefficients so that we an then use the other properties to combine them into a single logarithm:
{{{log(((5x+3)^4))-log(((x+15)^2))-log(((5x+2)^(1/2)))}}}
Since an exponent of 1/2 means square root, we can rewrite the last logarithm with a square root:
{{{log(((5x+3)^4))-log(((x+15)^2))-log((sqrt(5x+2)))}}}
Now we can use the first two properties. The difference in those two is that the first one has a "+" between the two logarithms and the second property has a "-". Since there is a "-" between the first two logarithms in your expression we will use the second property to combine them:
{{{log(((5x+3)^4/(x+15)^2))-log((sqrt(5x+2)))}}}
The remaining logarithms have a "-" between them so we will use the second property again to combine them:
{{{log(((((5x+3)^4/(x+15)^2))/sqrt(5x+2)))}}}
This may be an acceptable answer because it is, after all, a single logarithm. But the argument is a bit of a mess. One does not usually leave expressions with either fractions within fractions or with square roots in denominators. The following steps will "clean up" the argument into a better form:
{{{log((((5x+3)^4/(x+15)^2)*(1/sqrt(5x+2))))}}} Multiplying by the reciprocal is the same as dividing:
{{{log(((5x+3)^4/((x+15)^2*sqrt(5x+2))))}}}
{{{log((((5x+3)^4/((x+15)^2*sqrt(5x+2)))*(sqrt(5x+2)/sqrt(5x+2))))}}}
{{{log((((5x+3)^4*sqrt(5x+2))/((x+15)^2*(5x+2))))}}}
Now we have a single fraction with no square roots in the denominator. This is a better form than the earlier expression. I doubt that your teacher wants you to multiply out the numerator and denominator.