Another Log/Exponent Problem

[Jason76]

\(\displaystyle log_4(\dfrac{1}{4}) = y\) The goal is to find \(\displaystyle y\)

\(\displaystyle 4^{log_4(\frac{1}{4})} = 4^{y}\) By setting up everything as an exponent of 4, we can get rid of the \(\displaystyle log\) expression.

\(\displaystyle \dfrac{1}{4} = 4^{y}\) - This is a bit troublesome. In order to get \(\displaystyle y\) by itself, you have to take the \(\displaystyle log 4\) of both sides, but that just makes you go in a circle (you just go back from where you came).

So some people said to do this:

\(\displaystyle 4^{-1} = y\) - What happened on the right side?

 

[mmm4444bot]

\(\displaystyle log_4(\dfrac{1}{4}) = y\) The goal is to find \(\displaystyle y\)

So some people said to do this:

\(\displaystyle 4^{-1} = y\)

Those people are wrong because y = -1

log_4(1/4) is the exponent to which 4 must be raised so that the power equals 1/4. In other words, y is an exponent.

Use the property of exponents that tells you: 4^(-1) = 1/4

You can do this exercise in your head, if you memorize and understand the basic properties of exponents and the meaning of "log".

 

[Jason76]

Those people are wrong because y = -1

log_4(1/4) is the exponent to which 4 must be raised so that the power equals 1/4. In other words, y is an exponent.

Use the property of exponents that tells you: 4^(-1) = 1/4

You can do this exercise in your head, if you memorize and understand the basic properties of exponents and the meaning of "log".

In this case, the answer is there in front of you: \(\displaystyle y =\). You just the \(\displaystyle log_4\) of \(\displaystyle \dfrac{1}{4}\)