Simplifying Logarithms

Question

Simplify: w^logw(yz)

Answer 1

Kenneth is 100% correct, but that solution may not be satisfying your hunger to understand *why* that is true.

 

Let's say that your given exponential is equal to x

 

w^log_w(yz)= x

 

How do we normally deal with finding the solutions to exponentials? Well, we re-write them as logarithms. Recall that the rule for changing exponentials to logs is:

 

log_a (b) = x   <=>    a^x = b

 

So, let's do that here.

 

We use the base of the exponential (w) as our new base of a logarithm, placing the x as the new argument of the log. Then equate it to the exponent of the original exponential.

 

log_w (x) = log _w (yz)

 

To make this equation true, the arguments of each log need to be the same thing, so...

 

x = yz

 

Recall that we originally set your given (weird) exponential equal to x. So, that exponential is equal to yz, just like Kenneth said.

 

I hope this explanation helps you to remember *why* the rule that Kenneth said works. Also, remember that even if you get stuck on any problem, you can usually work through it with the tools that you have (like the conversion between logs and exponentials).

 

Best of luck in the future!

 

Answer 2

Simplify: w^logw(yz)

 

Observe: exponentiation of w (base)  to a logarithm, base w; the acts of exponentiating and log-taking are inverses, so they nullify one another, leaving the answer yz.