Question 818477
I'm assuming that the numbers immediately after "log" are supposed to be the bases of these logs. If I am wrong, then you will have to re-post your question. In the future, either use some English to describe your logs (like "base 5 log of (25) for {{{log(5, (25))}}}, or teach yourself algebra.com's syntax for formulas. (Click on the "Show source" link above to see what I typed to get that logarithm to display so nicely.)<br>
{{{21log(3, (root(3, x))) + log(3, (9x^2)) - log(5, (25))}}}
The following properties of logarithms are often used in problems like this:<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 are used to combine logs. They require that the logs have the same bases and coefficients of 1. The third property, since it allows us to "move" a coefficient out of the way, is often used to get coefficients of 1 so the other two properties may be used.<br>
The first log has a coefficient that is not a 1, it is 21. So we will use the third property to move the 21 into the argument as its exponent:
{{{log(3, ((root(3, x))^21)) + log(3, (9x^2)) - log(5, (25))}}}
We can use fractional exponents to simplify the first argument. Since a cube root is the same as an exponent of 1/3:
{{{log(3, ((x^(1/3))^21)) + log(3, (9x^2)) - log(5, (25))}}}
The rule for powers of a power is to multiply. Since 1/3*21 = 7:
{{{log(3, (x^7)) + log(3, (9x^2)) - log(5, (25))}}}<br>
Now we can use the first property to combine the first two logs:
{{{log(3, ((x^7)*(9x^2))) - log(5, (25))}}}
which simplifies to:
{{{log(3, (9x^9)) - log(5, (25))}}}<br>
We cannot use the second property to combine the remaining logs because the bases are different, 3 and 5. But the second log is one we can figure out "by hand". {{{log(5, (25))}}} represents the exponent one would put on a 5 to get a result of 25. We know what this exponent is. It is 2. So {{{log(5, (25)) = 2}}}. This makes our expression:
{{{log(3, (9x^9)) - 2}}}
This might be an acceptable answer. It is an expression with a single logarithm. But the problem, as you posted it, says "express as a single log" not "express with a single log" like we have above. If the final answer is really supposed to be just a single log and nothing else, then we have some more work to do.<br>
Here we can use a little trick. We are going to turn the "2" into a logarithm with a base of 3. This will allow us to combine the remaining terms into a single logarithm. {{{log(3, (3)) = 1}}} since the exponent for 3 that results in a 3 is 1. So I will multiply the 2 by (a weird-looking) 1:
{{{log(3, (9x^9)) - 2*log(3, (3))}}}
Now we use the third property to get the coefficient of 2 out of the way:
{{{log(3, (9x^9)) - log(3, ((3)^2))}}}
which simplifies to:
{{{log(3, (9x^9)) - log(3, (9))}}}
Now we can use the second property to combine the terms:
{{{log(3, ((9x^9)/9))}}}
The factors of 9 cancel:
{{{log(3, (x^9))}}}