Find the value of y.

log 4 64 = y

Question

Find the value of y.

log 4 64 = y

rock_mit182 · Answer

did you mean ?$$\log _{4} 64 =y$$

rock_mit182 · Answer

$$\log _{4}64= y \rightarrow 4^{y} =64$$

rock_mit182 · Answer

this is equivalent so you have to find out the exponent

solomonzelman · Answer

$\Huge\color{blue}{ \sf     log_464=y }$

$\large\color{red}{ Rule:~~\sf     log_ab^c=c~log_ab}$

$\Huge\color{blue}{ \sf     log_464=y }$

$\Huge\color{blue}{ \sf     log_44^3=y }$

$\Huge\color{blue}{ \sf     3~log_44=y }$

$\large\color{red}{ Rule:~~\sf     log_vv=1}$ (unless v=1, 0 , negative or complex) 

$\Huge\color{blue}{ \sf     3~log_44=y }$

$\Huge\color{blue}{ \sf     3\times 1=y }$

$\Huge\color{blue}{ \sf     y=3 }$

anonymous · Answer

yea n thanks i have the answer already @rock_mit182 but can u help me with another question im stuck on well im stuck on like three of them

mathmale · Answer

One method would be to take advantage of the inverse property of the log and exponential functions.   y=log (to the base 4) of x and y=4^x are inverses of one another.

Thus, raising 4 to the power (log to the base 4 of) x results in simply x:

$$4^{\log_{4}x }=x$$

So, given      4             =             4,

raise the first 4 to the power $$\log_{4} 64$$ and the other 4 to the power y.

anonymous · Answer

The function y = x2 is transformed to y = (x + 4)2. Which statement is true about the transformed function? 
a.It is an even function.
b.It is an odd function.
c.It is neither an even nor an odd function.
d.It is both an even and an odd function.

mathmale · Answer

Another way would be to recognize that 64 is 4^3.  Therefore,

$$y=\log_{4} 64=\log_{4} 4^3 =3$$

rock_mit182 · Answer

you have to test this fact by using the definitions of an even and odd

If a function f satisfies f(-x)=f(x) for every number x in its domain, then f is called
an even function. For instance$$f(x)=x ^{2}$$, the function is even because
$$f(-x) =(-x)^{2} =(-1)^{2}(x)^{2} =(x)^{2}=f(x)$$

mathmale · Answer

Your posting two separate questions in one conversation is confusing, as you can see from my postings in the middle of your second question.  Would you please post each question separately from now on.  Thank you.

anonymous · Answer

@mathmale okay i will i already had the answer to the first one