Question

HELLOOOO HELP PLEASE

Answer 1

Answer:

$2*log(x)+log(y)$

Step-by-step explanation:

So, there are two logarithmic identities you're going to need to know.

Logarithm of a power:

   $log_ba^c=c*log_ba$

   So to provide a quick proof and intuition as to why this works, let's consider the following logarithm: $log_ba=x\implies b^x=a$

   Now if we raise both sides to the power of c, we get the following equation: $(b^x)^c=a^c$

   Using the exponential identity: $(x^a)^c=x^{a*c}$

    We get the equation: $b^{xc}=a^c$

    If we convert this back into logarithmic form we get: $log_ba^c=x*c$

    Since x was the basic logarithm we started with, we substitute it back in, to get the equation: $log_ba^c=c*log_ba$

Now the second logarithmic property you need to know is

The Logarithm of a Product:

    $log_b{ac}=log_ba+log_bc$

    Now for a quick proof, let's just say: $x=log_ba\text{ and }y=log_bc$

    Now rewriting them both in exponential form, we get the equations:

    $b^x=a\\b^y=c$

    We can multiply a * c, and since b^x = a, and b^y = c, we can substitute that in for a * c, to get the following equation:

    $b^x*b^y=a*c$

   Using the exponential identity: $x^{a}*x^b=x^{a+b}$, we can rewrite the equation as:

 

   $b^{x+y}=ac$

   taking the logarithm of both sides, we get:

   $log_bac=x+y$

   Since x and y are just the logarithms we started with, we can substitute them back in to get: $log_bac=log_ba+log_bc$

Now let's use these identities to rewrite the equation you gave

$log(x^2y)$

As you can see, this is a log of products, so we can separate it into two logarithms (with the same base)

$log(x^2)+log(y)$

Now using the logarithm of a power to rewrite the log(x^2) we get:

$2*log(x)+log(y)$