Question Video: Evaluating Logarithms Mathematics

What is the value of log_(3) 81√(3)?

02:59

Video Transcript

What is the value of log base three of 81 times the square root of three?

To solve this problem, we’re gonna have to use some of our logarithm rules to simplify the expression. The first thing we notice is inside the log, we’re multiplying 81 by the square root of three. We can take that and expand it into two separate expressions, which would be the log base three of 81 plus the log base three of the square root of three. And because we are working with the base three, one way to simplify would be to see if we could write 81 and the square root of three with a base of three and some exponent.

We know that 81 is nine times nine and that nine is three squared. And that means we can rewrite 81 as three to the fourth power. We now have log base three of three to the fourth power. And in a similar way, we can rewrite the square root of three as three to the one-half power. We also know that log base 𝑏 of 𝑥 to the 𝑛 power is equal to 𝑛 times log base 𝑏 of the 𝑥. And that means we can rewrite the first part as four times log base three of three and the second part as one-half times log base three of three. When we’re taking log with a base 𝑏 of 𝑏, it equals one. So the log base three of three equals one.

And the first term will be four times one. Again, we have log base three of three, which is one, making our second term one-half times one. We end up with four plus one-half. To add the fractions, we wanna a common denominator. So we have eight over two plus one over two, which equals nine-halves. What I have shown here is probably the most common way to solve the problem. But I wanna show one other method. Instead of breaking this log up into two terms, we could’ve initially said that 81 equals three to the fourth power. And the square root of three equals three to the one-half power, which means 81 times the square root of three is equal to three to the fourth power times three to the one-half power.

Based on what we know about exponents, when we multiply exponents with the same base, we add their power values, which would be three to the four plus one-half power. We’ve already found that four plus one-half is nine-halves. We then could’ve substituted three to the nine-halves power back into our log and say that log base three of three to the nine-halves power is equal to nine-halves times the log base three of three, which is simply nine-halves. I do think it’s probably more common to do the expanded form. However, this is another option.

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