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Solving Simple Logarithmic Equations - Problem 2
Solving a simple logarithmic equation. So whenever we're solving a logarithmic equation, what we want to do is put it into exponential form.
For this particular example, we're missing one of our components of logarithmic form. We're used to having this equal to something, and what I want to bring to your attention is that when you're used to solving problems that don't necessarily have that component. If I say 2 plus 3, you know by default you can put these together and this is going to be equal to 5.
By same logic anything I say 3², even without an equal sign, you know that this is equal to 9. You know that these are both equal to something, so what you can really do in order to help you through a logarithmic equation like this, is to say okay this is equal to something, we don't know what something is, so we can call it x.
Now we have our components of logarithmic form, put into exponential form. So the cube root of 7 stays on one side and we end up with 49 to the x. Now we have turned this into a exponential problem. We have two things, different bases but we need to somehow rewrite these bases as the same.
The cube root is the same thing as the to the power of 1/3, so this is 7 to the 1/3, so we need to somehow write 49 as the same base, 49 is 7², so this becomes 7² to the x. Power to a power you multiply, 1/3 is equal to 7 to the 2x. When our bases are the same, our exponents have to be the same, ending up with 1/3 is equals to 2x, divide by 2 multiply by 1/2 and we end up with one-sixth is equal to x.
X is the thing we introduce to this problem so it's really not part of our answer, so our answer is just going to be 1/6.
So whenever you're dealing with a logarithmic equation that you are missing a term, just put equals x. This is equal to something, x can help us out and then for this particular example we have to go back to the rules of exponents, solve it out in order to get the bases the same and set our exponents equal. Solving a logarithmic equation.