Solve each equation.

Question

x = \log_{4} \sqrt[3]{16}

Accepted Answer

Hey, everyone in this problem, we're asked to solve the following logarithmic equation. We have that X is equal to log based 11 of the seventh group of 121. We're given four answer choices. All of them are sets containing just a single value for a, that value is 2/7 for B, it's 1/7 for C, it's seven divided by 11. And for D it's two, we're gonna start by rewriting what we were given. We have X is equal to a log base of the seventh roof of 121. Now, we can tackle this problem a couple of different ways. The first way we could write this equation as an exponential equation. OK. We have a rule that allows us to write a log equation as an exponential equation. So we can convert this to an exponential equation and simplify and work through it. That way the other thing we can do, which is how I'm gonna approach the problem. We already have this as X equals something. So let's just try to simplify that right hand side. OK? And again, either of these approaches work and they're about the same length. So whichever way you like to approach this problem. Now, the first thing we're gonna do is we're gonna rewrite this route inside. And we know that we have a log rule that allows us to deal with exponents inside of our log. So we want to write this route as an exponent so that we can use that. Now recall if we have the N root of some value, say A, we can write this as a to the exponent one divided by N. So our equation becomes X is equal to log phase of 121 to the exponent 17. Now we have an exponent inside of our log. I mentioned that we have a rule for this. So let's recall that if we have log thanks of something, say A to the exponent N, we can write this as N multiplied by log base B of A. So we have an exponent inside of our log. We can bring it out and multiply it in front. Applying that to our problem. We have the exponent 1/7 that's gonna come out front and multiply our lock. So now we have 1/7 log base of 121 and that exponent is gone from inside of a log. Now, OK. We're getting close. Now we're down to log base 11 of a number 121. Now let's call if we have log base B A B, this is just equal to one. OK. So if we can get this, as long as 11 of 11, we know that it's gonna be equal to one. Well, we know that 100 and 21 is equal to 11 squared. So let's write it like that. We have that X is equal to 1/7 multiplied by log base 11 of 11 square. Now we have another exponent inside of our log. We can use the exact same rule that we used with our exponent of 1/7 to bring that down. And in front, we get that X is equal to two sevens on base 11 of 11. Now we have log base 11 of OK, the base and what's inside the log are the same and this is just going to be equal to one. And so we get the X is equal to 2/7 multiplied by one which we know is just going to be equal to two sevens. And so the solution to this equation we were given is X is equal to two sevens which corresponds with answer choice. A thanks everyone for watching. I hope this video helped see you in the next one.

Solve each equation. $x = \log_{4} \sqrt[3]{16}$

Key Concepts

Logarithms and Their Properties

Radicals and Rational Exponents

Change of Base and Simplification Techniques

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